 How Many 9s?How many 9s are in the decimal expansion of 99999989999^{2}? Solution 9. The given number is 10^{11} − 10001. Its square is 10^{11}(10^{11} − 20002) + 100020001. The number of 9s in this equals the number of 9s in 10^{11} − 20002 = 99999979998.
From May Calendar, May 2007

 49 Consecutive IntegersThe sum of 49 consecutive integers is 7^{5}. What is their median? Solution 343. The sum of a set of integers is the product of the mean and the number of integers, and the median of a set of consecutive integers is the same as the mean. So the median (according to the definition of mean) must be 7^{5} divided by 49; which is 7^{5} divided by 7^{2}; which can be simplied to 7^{3}, or 343.
From March Calendar, Mathematics Teacher, September 2005

 Ten Miles of PenniesApproximately how much is a tenmilehigh stack of pennies worth? Solution $92160. According to our estimates, a penny is 0.06785 inches thick; twenty pennies stand 1 3/8 inches tall. Since ten miles is 5280⋅12⋅10, or 633600, inches, a tenmilehigh stack of pennies would contain 633600/0.06875, or 9216000, pennies. It would be worth $92160.
From March Calendar, Mathematics Teacher, March 2001

 Divisors of 80!What is the smallest natural number that is not a divisor of 80!? Solution 83. Certainly 1, 2, 3, ..., and 80 will all be divisors; 81 = 3 × 27 and 82 = 2 × 41 will also be divisors. However, 83 is prime and will not be a divisor.
From January Calendar, Mathematics Teacher, January 1999

 DiagonalsHow many diagonals does a convex polygon with twenty sides have? Solution 170. A diagonal connects a vertex of the polygon to a nonadjacent vertex, so seventeen diagonals extend from each vertex. Since each diagonal gets "counted" twice, that is, once for each vertex, the number of diagonals is 20(17)/2, or 170.
From March Calendar, Mathematics Teacher, March 2002

 Distinct PrimesThe sum of three distinct primes is 40. What is their product? Solution
434. All primes except 2 are odd, and the sum of any three odd numbers is odd. For the sum of three primes to be even, one of them must be 2. This implies that the sum of the remaining two primes is 38. Of the two possible solutions (7 + 31 = 38 = 19 + 19), only one yields a set of distinct primes. Thus, we seek 2 + 7 + 31 = 40 and 2 × 7 × 31 = 434.
From Calendar Problems , December 2008

 Common MultipleWhat is the least common multiple of the first ten counting numbers? Solution
2520. One easy way to solve this problem is to consider the prime factorization of the first ten counting numbers, starting with the largest value and working down to the smallest: 10 = 5 • 2; 9 = 3^{2}; 8 = 2^{3}; and 7. All remaining counting numbers (6, 5, 4, 3, 2, and 1) can be constructed by using the factors of the preceding (1 is a factor of all). The least common multiple is therefore 5 • 3^{2} • 2^{3} • 7 = 2520.
From Calendar Problems , November 2008

 7 as a DigitHow many twodigit positive integers have at least one 7 as a digit? Solution 18. There are 10 twodigit numbers with a 7 as the 10s digit, and 9 twodigit numbers with 7 as the units digit. Because 77 satisfies both of these properties, the answer is 10 + 9 – 1 = 18.
From March Calendar, Mathematics Teacher, August 2005

 3rd Base Multiplication Assuming that both numbers are in base 3, find the product of 21⋅21. Solution
1211. One way to proceed is to convert from a base3 representation (n_{3}) to a base10 representation (n_{10}), perform the multiplication, and then convert back to base 3. This leads to 21_{3} = 2⋅3^{1} + 1⋅3^{0} = 7_{10}, or just 7. Thus, 21_{3} × 21_{3} = 7 × 7 = 49, and 49 = 1⋅3^{3} + 2⋅3^{2} + 1⋅3^{1} + 1⋅3^{0} = 1211.
A second method uses the traditional algorithm and notation in base 3:
21
× 21
21
112
1211
From April Calendar, Mathematics Teacher, April 2008

 13 FactorsWhat is the least positive integer that has exactly thirteen factors? Solution
4096. The answer must be a square number, since it has an odd number of distinct factors. The prime factorization of this number will be 2^{13 − 1} = 2^{12} = 4096.
Alternatively, the following is a more rigorous solution to the problem. The number of factors of an integer is given by (n_{1} + 1)⋅(n_{2} + 1)⋅(n_{3} + 1)..., where the prime factorization of the number is
P_{1}^{n1}⋅P_{2}^{n2}⋅P_{3}^{n3}⋅...
Since 13 is prime, the prime factorization must be of the form k^{12}. Since 2 is the least possible value for k, the least possible number is 2^{12}, or 4096.
From October Calendar, Mathematics Teacher, October 1999

 Only EvensHow many threedigit numbers consist only of even digits? Solution 100. The number of threedigit numbers consisting of only even digits is not 125. Although the second and third digits can be chosen from any of 0, 2, 4, 6, and 8, the first digit cannot be 0, so 4 × 5 × 5, or 100, threedigit numbers have only even digits.
From October Calendar, Mathematics Teacher, October 2001

 Tens DigitWhat is the tens digit of 0! + 1! + 2! + 3! + ... +2000!? Solution
1. The first ten terms of the sequence are these:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
Adding the last two digits of the first ten terms result in 214. Since 10!, or 3,628,800, ends in two zeros, all subsequent terms of the sequence will end in at least two zeros because 10! will be a factor of each. The tens digit of the sum of the terms of the sequence is thus 1.
From October Calendar, Mathematics Teacher, October 1999

 SecondsOn average, how many seconds does the month of February have? Solution
2,440,152 seconds. February has twentyeight days, except during a leap year, when the month has twentynine days. A leap year occurs in a year that is divisible by 4 unless the year is a divisible by 100 but not 400. That explanation gives the reason that 2000 was a leap year even though it was divisible by 100. Therefore, in a 400year span, ninetyseven leap years occur; February has an average of 28.2425 days. Since 3,600 seconds are in an hour, 3,600 × 24, or 86,400, seconds are in a day; so 86,400 × 28.2425, or 2,440,152, seconds are in an average February.
From February Calendar, Mathematics Teacher, February 2002

 Operation If a*b = a^{b} – b, find (2*3)*4. Solution
621. (2*3)*4 = (2^{3} – 3)*4 = (8 – 3)*4 = 5*4 = 5^{4} – 4 = 625 – 4 = 621.
From March Calendar, Mathematics Teacher, April 2005

 111,111,111^{2} Find the exact value of (111,111,111)^{2}. Solution
12,345,678,987,654,321. This solution can be found using the traditional algorithm for multiplication. Note that there will be no carrying involved.
Alternate solution. Observe that 11^{2} = 121; 111^{2} = 12,321; 1111^{2} = 1,234,321 and proceed accordingly.
From February Calendar, Mathematics Teacher, February 2008

 DigitsRamona just completed writing her first book and numbered the pages. Numbering the book required 2649 digits. How many pages does Ramona’s book have? Solution 919 pages. The numbering for the first 9 pages uses 9 digits, and the numbering for pages 10–99 (90 pages) uses 180 digits. So far, 189 digits out of the 2649 have been used, leaving 2460 digits. Dividing 2460 by 3 gives 820. Ramona’s book had 820 pages with 3digit numbers, 90 pages with 2digit numbers, and 9 pages with 1digit numbers. This gives 919 pages and a total of 2649 digits.
From Calendar Problems , September 2008

 Pascal's LittlestWhat is the smallest 4digit number to appear in the row of Pascal’s triangle in which the first 4digit number appears? Solution 1287. See the Mathematics Teacher journal from December 1999 for a detailed solution.
From December Calendar, Mathematics Teacher, December 1999

 Divisible by 3, 4, and 5How many of the first 500 positive integers are divisible by 3, 4, and 5? Solution Eight.
Since 3, 4, and 5 are relatively prime, their least common multiple is their
product, 60. Only multiples of 60 can be divided by 3, 4, and 5. Eight multiples of 60 are less than 500.
From September Calendar, September 2003

 Only OddsHow many threedigit numbers consist only of odd digits? Solution 125. Each of the three digits can be any of the five odd digits 1, 3, 5, 7, and 9. A total of 5*5*5; or 125, such threedigit numbers exist.
From October Calendar, Mathematics Teacher, October 2001

 poster_problem62Find a Pythagorean triple where each integer is odd. Solution There is none. If each leg is an odd integer, their squares are odd integers and the sum of the squares would be even.

 Choosing While BlindfoldedA bowl contains 50 colored balls: 13 green, 10 red, 9 blue, 8 yellow, 6 black,
and 4 white. If you are blindfolded, what is the smallest number of balls you must
pick to guarantee that you have at least 7 balls of the same color?
Solution 35.
You might choose 6 greens, 6 reds, 6 blues, 6 yellows, 6 blacks, and 4
whites. The next one will certainly give you 7 of some color.
From March Calendar, March 2006

 Perfect SquaresWhich perfect squares are in the arithmetic sequence {2, 8, 14,...}? Solution No perfect squares occur in this sequence, since for all integers, n, n^2 is never 2 or more than a multiple of 6.
See page 760 of the December 2001 issue of MT for two methods of confirming this result.
From December Calendar, Mathematics Teacher, December 2001

 Prime FactorWhat is the largest prime factor of 8091? Solution 31.
Write 8091 as 8100 minus 9.
Recognize the difference in squares and simplify.
Then, 8091 has factors 3, 3, 29, and 31.
So 31 is its largest prime factor.
From May Calendar, May 2007

 Probability of HeadsTwo fair coins are flipped simultaneously. This is done repeatedly until at least one
of the coins comes up heads, at which point the process stops. What is the probability
that both coins came up heads on the last flip?
Solution 1/3.
The previous flips are irrelevant. Of the four equally likely outcomes of a double
flip, HH, HT, TH, and TT, the first three have at least one head, and just
one of them has the other coin also a head.
From March Calendar, March 2006

 Partitioning a SquareA square is partitioned into 30 nonoverlapping triangles so that each of the four sides of the square is also a side of one of the 30 triangles. The intersection of any two triangles is empty, a common vertex, or a common edge. How many points in the interior of the square serve as vertices of one or more triangles?
Solution 14.
If there are v interior vertices, then the total number of degrees in all
the angles (including the interior vertices and the four vertices of the square)
is, on the one hand, 30(180) and, on the other hand, 360v + 4 (90). Thus, 30 =
2v + 2, and v = 14.
From March Calendar, March 2006

 One Red Face A cube has an edge length of 4 in. The cube is painted red all over and then cut into 64 cubes of edge length 1 in. How many of these cubes have exactly one face painted red? Solution
24. When the original cube is divided into 64 smaller cubes as shown, each face has only 4 cubes in the center with exactly 1 face painted red. (The cubes on the interior of the original cube will have no painted faces, and the other exterior cubes will have at least 2 painted faces.) This will result in 6 faces with 4 cubes each, or 4 × 6 = 24 cubes with exactly 1 face painted red.
From April Calendar, April 2007

 Rainy DaysIt rained on exactly 7 of the days during Jane's trip. On each day that it rained, it rained either in the morning or in the afternoon but not both. There were exactly 5 afternoons when it did not rain and exactly 6 mornings when it did not rain. How many days did the trip last? Solution
9. Let M be the number of days it
rained in the morning; let A be the number of days it rained in the afternoon; and let N be the number of days when it did not rain. We have M + A = 7 (days when it rained); M + N = 5 (afternoons when it did not rain); and A + N = 6 (mornings when it did not rain). Adding the equations we get 2(M + A + N) = 18, M + A + N = 9.
From August Calendar, August 2007

 Chicken NuggetsChicken nuggets come in packets of 6, 9, or 20. What is the largest number of nuggets that you cannot buy when combining various packets?
Solution 43.
Any multiple of 3 that is greater than 3 can be obtained from packets of 6 and 9 nuggets. Since 36 = 9 + 9 + 9 + 9, 38 = 20 + 9 + 9, and 40 = 20 + 20, any even number ≥ 36 can be achieved by adding 6s to each of these. Similarly, by adding another 9, any odd number ≥ 45 can be achieved. But 43 is not yet guaranteed, so we need to examine the possibility of combinations that yield 43
nuggets. Since 43 is not a multiple of 3, likewise, 43 – 20 = 23 is not a multiple
of 3, and 43 – 2 • 20 = 3 is too small to achieve. Consequently, 43 cannot be
obtained.
From March Calendar, March 2006

 Subtracting Lengths? Jim and Bob were both asked to solve the following:
43'7" − 28'11"
Jim answered 14'8"; Bob's answer was 14'56". Who is correct? Solution Jim and Bob can both be correct, depending on the units used. Jim was using feet (') and inches ("), while Bob was using angle (or time) units of minutes (') and seconds (").
From August Calendar, August 2007

 Path DistanceJeremy walks along a spiral path (as shown). If the path is 2 meters wide, how far does he walk?
Solution 97 meters. By superimposing a grid of 2X2 squares, it is easy to see that the path Jeremy travels can be broken into 13 sections of lengths 13, 12, 12, 10, 10, 8, 8, 6, 6, 4, 4, 2, and 2 meters.
From August Calendar, August 2007

 Pluto Math Pluto's inhabitants use the same mathematical operators that we do (+, −, etc.). They also use and operator, @, that we do not know. The following are true for any real numbers x and y.
x @ 0 = x
x @ y = y @ x
(x + 1) @ y = (x @ y) + y + 1
What is the value of 12 @ 5? Solution
77. We have
12 @ 5 = 5 @ 12
= (4 @ 12) + 13
= (3 @ 12) + 13 + 13
= (3 @ 12) + 26.
Continuing,
(3 @ 12) + 26 = (2 @ 12 + 39)
= (1 @ 12) + 52
= (0 @ 12) + 65
= (12 @ 0) + 65
= 77.
From August Calendar, August 2007

 10 Digit NumberWrite a tendigit number so that the first digit indicates how many 0s are in the number, the second digit indicates how many 1s are in the number, the third digit indicates the number of 2s, etc. Solution 6,210,001,000.
From August Calendar, August 2007

 Flat TireAfter a cyclist has gone 2/3 of his route, he
gets a flat tire. Finishing on foot, he spends
twice as long walking as he did riding. If his
walking and riding rates are both constant, how much faster does he ride than walk? Solution 4 times as fast. He walks onethird of the way, or half as far as he rides, but it takes him twice as long. Therefore, he rides four times as fast as he walks.
From May Calendar, May 2007

 Rectangular SolidsHow many rectangular solids are possible with a volume of 100 cubic meters and sides of integral dimensions? Solution 8. The solutions by dimension are (1, 1, 100), (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10), (2, 2, 25), (2, 5, 10), and (4, 5, 5).
From August Calendar, August 2007

 Milk or Bread or Both?A customer enters a supermarket. The probability that the customer buys bread is .6, the probability that he buys milk is .5, and that he buys both bread and milk is .3. What is the probability that the customer would buy either bread or milk or both? Solution
.8, or 4/5. Let B represent the event that the customer buys bread, M the event that the customer buys milk. Then, according to the rule of addition, we have
P(B ∪ M) = P(B) + P(M) − P(B ∩ M)
= .60 + .50 − .30
= .80
Alternate solution. A Venn diagram offers a visual approach. If we first show the probability that the customer buys mile and bread—P(M ∩ B)—then we can complete the diagram by subtraction: P(M ∪ B) = .2 + .3 + .3 = .8
From May Calendar, May 2007

 Engine FailureAn aircraft is equipped with three engines that operate independently. The probability of an engine failure is .01. What is the probability of a successful flight if only one engine is needed for the successful operation of the aircraft? Solution
.999999. Let P(S) be the probability of a successful flight, P(S') the probability of an unsuccessful flight, and P(F_{n}) the
probability of n engines failing. Since the flight is unsuccessful only when all three
engines fail, then the probability of unsuccessful flight is
P(S') = P(F_{1} ∩ (F_{2} ∩ F_{3}))
=(.01)(.01)(.01)
=(.01)^{3}.
But
P(S) = 1 − P(S')
= 1 − (.01)^{3}
= 1 − .000001
= .999999.
From May Calendar, May 2007

 Equal Row SumsUsing the diagram, place the numbers 1
to 10 in the circles so that the sums in the
rows of three circles are the same and the
sums in the rows of four circles are the
same.
Solution One possible arrangement is shown. The sum of the rows of four is 23, and the sum of the rows of three is 16.
From August Calendar, August 2007

 3D ShadowsA threedimensional figure is formed so that the diagrams shown represent three shadows cast. Describe a possible figure.
Solution The figure could be a circular cylinder with a diagonal slice removed, as shown. This would allow a top shadow view of a circle. From the "left" or "right," it would have a square shadow, and from the "front," it would have a
triangular shadow.
From August Calendar, August 2007

 Palindromic SpeedingThe odometer of a family car shows 15,951 miles. The driver noticed that this number is palindromic: it reads the same backward as forward. Surprised, the driver saw his third palindromic odometer reading (not counting 15,951) exactly five hours later. How many miles per hour was the car traveling in those 5 hours (assuming speed was constant)? Solution 62. Realistically, the first digit of 15,951 could not change in 5 hours. Therefore, 1 is the first and last digit of the new number. Clearly, the second and fourth digits changed to 6. So the next three resulting palindromes are 16,061, 16,161, and 16,261. Thus, the car traveled 310 miles in 5 hours and must have been traveling 62 miles per hour.
From May Calendar, May 2007

 WorkshopToni makes stools and tables in her workshop.
At the end of a long day, she has made 35 legs. If each stool has 3 legs and each table has 4 legs, and the legs are all identical, how many stools and tables can she now assemble? Solution 1 stool and 8 tables; or 5 stools and 5 tables; or 9 stools and 2 tables. Since each stool requires 3 legs and each table requires 4 legs, the scenario can be described as all possible integral solutions for 3x + 4y = 35. This has 3 possible solutions: (1, 8), (5, 5), (9, 2). Note that in terms of legs, 4 stools are equivalent to 3 tables.
From August Calendar, August 2007

 Poor Fly Solution
From May Calendar, May 2007

 HexagonLew is playing darts on a starshaped dartboard
in which two equilateral triangles trisect the sides of each other as shown. Assuming that a dart hits the board, what is the probability that it will land inside the hexagon?
Solution 1/2. As shown, each triangle can be reflected to the interior of the hexagon in such a way that the triangle areas are equal to the area of the hexagon. In this manner, the area of the hexagon is half that of the entire dartboard.
From August Calendar, August 2007

 Shortest PathWhat is the length of the shortest path that
can connect the twentyfive dots shown?
The vertical and horizontal distance between
any two adjacent dots is 1 unit.
Solution 24 units. There are several paths that can be taken of this length. One such example is shown. In this case: 5(4) + 4(1) = 24.
From August Calendar, August 2007

 World SeriesThe World Series is a series of games that continues until one team has won four games. The Braves and the Yankees are playing. Each team has a 1/2 probability of winning each game. Given that exactly six games are played, what is the probability that the Braves win the World Series? Solution 1/2.
Since both teams are equally likely
to win each game, they are equally likely to
win in six games. Therefore, given that the
Series has six games, the Braves have a
probability of 1/2 of winning the Series.
From May Calendar, May 2004 MT.
Submitted by Richard Rusczyk, cocreator of the Mandelbrot Competition and coauthor of The Art of Problem Solving, volumes 1 and 2.

 PencilsIf p pencils cost c cents, how many pencils can be purchased for d dollars? Solution 100 dp/c pencils.
If you can buy p pencils for c cents, then you can buy 100p pencils for c dollars (100c cents), so you can buy 100p/c pencils for 1 dollar. Therefore, d dollars is the cost of d(100p/c), or 100dp/c pencils.
From February Calendar, February 2004 MT.
From The Ideas of Algebra, K12, the 1988 NCTM Yearbook (Reston, Va.: National Council of Teachers of Mathematics, 1988).

 Fair DieA fair die is tossed four times. What is the probability that it lands with either 5 or 6 on top at least once? Solution Approximately .80247.
The number of possible ways to roll four dice is 6^{4}, or 1296: six choices for each of the four rolls.
There are four ways to roll a single die once and not get a 5 or 6 (that is, to get a 1, 2, 3, or 4), so the number of ways to roll a die four times and not get a 5 or a 6 is 4^{4}, or 256.
Then 1296  256, or 1040, ways (the rest of the possibilities) exist to roll the die four times and get a 5 or 6 at least once.
Therefore, the probability of rolling a die four times and getting a 5 or a 6 at least once is 1040/1296, or 65/81, which is approximately .80247.
From November Calendar, November 2003 MT.
From The Contest Problem Book I: Annual High School Mathematics Examinations 19501960, compiled and with solutions by Charles T. Salkind (Washington, D.C.: Mathematical Association of America, 1961).

 Trace the FigureCan you trace this figure without lifting your pencil? If no, why not? If yes, where must you start?
Solution Yes; you can start anywhere.
Since all the vertices of the diagram are of an even degree, tracing the figure is possible. Moreover, the diagram can be traced regardless of the starting point.
From November Calendar, November 1999 MT.
Contributed by Hugh Thompson, Hawken School, Gates Mills, Ohio.

 Earrings In a certain village live 800 women. Three percent of them are wearing one earring. Of the other 97 percent, half are wearing two earrings and half are wearing none. What is the total number of earrings being worn by the women? Solution
800. The number of women wearing two earrings and the number of women wearing no earrings are the same, so the average number of earrings per woman in this group is 1. Since the other women are wearing one earring each, the total number of earrings is equal to the total number of women, 800.
From October Calendar, Mathematics Teacher, October 1999

 A Probability ProblemThe probability that both Adrian and Brian answer a question correctly is 18 percent. The probability that one, but not both, of them answers the question correctly is 54 percent.
Given that Adrian and Brian are operating independently, what is the probability that Adrian answers the question correctly, given that Adrian is more likely than Brian to answer correctly?
Solution .6.
Let a and b denote the probabilities that Adrian and Brian, respectively, answer the question correctly. Then ab = .18 and a(1  b) + (1  a)b = .54. The second equation simplifies to a + b  2ab = .54.
Substituting ab = .18 gives us a + b  2(.18) = .54, so a + b = .9, which implies that a + 18/a = .9, so a^{2}  .9a + .18 = 0. Factoring the lefthand side of the equation reveals that (a  .6)(a  .3) = 0, so a is either .6 or .3, and b is either .3 or .6, respectively.
Since a > b, we conclude that the probability that Adrian gets the answer correct is .6.
From February Calendar, February 2001 MT.
Submitted by Richard Kalman, Executive Director of Mathematical Olympiads for Elementary and Middle Schools, Bellmore, N.Y.

 Dart Board A dart board is constructed with concentric circles of radii in the ratio 1:2:3 and congruent sectors as shown. If a dart is thrown and lands somewhere on the board randomly, what is the probability that it lands on a shaded section?
Solution
1/3. By reorganzing, it can easily be shown that the shaded region makes up onethird of the total area.
From December Calendar, Mathematics Teacher, December 2007

 Tom and BillTom is standing in a hole that is 4 feet deep. Bill asks him how much deeper he is going to dig the hole. Tom replies that he will dig 4 feet 2 inches deeper and that the top of his head will then be the same distance below ground level that it is now above ground level. How tall is Tom? Solution 6 feet 1 inch. The top of his head will go down 4 feet 2 inches with the additional digging. Half that distance was above the hole before the additional digging, so he is 4 feet + 2 feet 1 inch, or 6 feet 1 inch, tall.
From May Calendar, Mathematics Teacher, May 1999

 What Is the Price?Matthew and Matilda want to buy a set of DVDs. Matthew has $47 less than the purchase price, and Matilda has $2 less. If they pool their money, they still do not have enough to buy the DVDs. If the set costs an integral number of dollars, what is its price? Solution $48.
If P is the price of the DVDs, Matthew has P  47 dollars and Matilda has P  2 dollars.
We know that (P  47) + (P  2) < P, since the amount of their combined savings is still less than the price of the DVDs.
Solving for P, we have P < 49; but P > 47, since Matthew had some money. Therefore, P is a whole number such that 47 < P < 49, so the price of the DVDs is $48.
From December Calendar, December 2002 MT.
Adapted from 50 Mathematical Puzzles and Recreations, Orange Collection (Emeryville, Calif.: Key Curriculum Press).

 InvestmentsA man has $10,000 to invest. He invests $4,000 at 5 percent and $3,500 at 4 percent. To have a yearly income of $500 from the investment, at what rate must he invest the remainder of the money? Solution At 6.4 percent or higher.
The man wants to earn $500 in interest each year. Investing $4000 at 5 percent yields yearly interest of 0.05 x $4000, or $200; whereas $3500 invested at 4 percent yields 0.04 x $3500, or $140; so the remaining $2500 needs to be invested at a rate that yields interest of $160 per year (to have a total interest of $500 each year). If r is the annual percentage rate for the $2500, then
so the man needs to invest the remaining $2500 at 6.4 percent or higher to guarantee an interest income of at least $500 a year.
From November Calendar, November 2003 MT.
From The Contest Problem Book I: Annual High School Mathematics Examinations 19501960, compiled and with solutions by Charles T. Salkind (Washington, D.C.: Mathematical Association of America, 1961).

 FacetiousIn how many arrangements of the letters in the word FACETIOUS are the vowels in alphabetical order? Solution 3024 ways.
A total of 9! ways exist to arrange the letters in the word FACETIOUS (nine ways of picking the first letter exist, eight ways to pick the second exist, and so on).
For each arrangement of the consonants
in the word, the vowels can be arranged in 5!
ways. In only one of them are the vowels in
alphabetical order.
Hence, 1/5! of the arrangements have the vowels in alphabetical order, for a total of (9!/5!), or 3024.
From May Calendar, May 2004 MT.
Submitted by Richard Rusczyk, cocreator of the Mandelbrot Competition and coauthor of The Art of Problem Solving, volumes 1 and 2.

 Wake Up!During the first four days of Arthur's new job, he had to wake up at 5:30, 5:30, 7:10, and 7:30. On average, at what time did he have to wake up each morning? Solution 6:25.
We employ a useful technique for computing averages that is often faster and easier than simply adding the numbers and dividing by four. Instead, we tally the number of minutes away from 6:00 that Arthur awoke each morning. We obtain 30, 30, 70, and 90. Their average is
minutes. Therefore, on average, Arthur awoke 25 minutes after 6:00, or at 6:25.
From May Calendar, May 2003 MT.
Submitted by Sam Vandervelde, director of the Mandelbrot Competition.

 How Tall Was the Tree?Lightning hit a tree onefourth of the distance up the trunk from the ground and broke the tree so that its top landed 60 feet from its base, as shown. How tall was the tree originally?
Solution Approximately 84.85 feet tall.
Let h be the height, in feet, of the tree before it fell. Then (h/4)^{2} + 60^{2} = (3h/4)^{2}, so h^{2}/16 + 3600 = 9h^{2}/16.
Solving for h reveals that h^{2}/2 = 3600, so h = , and the original height of the tree equals , or approximately 84.85 feet.
From March Calendar, March 2003 MT.
From the 19911992 MATHCOUNTS national competition.

 How Many Cats Does Catwoman Have?If you ask Batman's nemesis, Catwoman, how many cats she has, she answers with a riddle: "Fivesixths of my cats plus seven." How many cats does Catwoman have? Solution Fortytwo cats.
If the number of cats is seven more than fivesixths the number of cats, then seven must be onesixth of the number of cats. That is, Catwoman had 7 x 6, or fortytwo, cats.
From April Calendar, April 2003 MT.
From Mathematical Games, by Marie Berrondo (Englewood Cliffs, N.J.: Prentice Hall, 1983).

 etraders Three students open etrade accounts and become day traders. Although they all work hard, they achieve the following steady rates of losing money: The first student loses $1000 in one hour, the second student loses $1000 in two hours, and the third student loses $1000 in three hours. Find the number of minutes it takes for the three students together to lose a total of $2000. Solution
About 65 minutes (to the nearest minute). The losing rates are $1000/hr., $500/hr., and $333.33/hr. The combined losing rate is $1833.33/hr. Thus,
From Calendar Problems , September 2008

 Faulty Timepieces Grandma’s watch gains 30 minutes every hour, while Grandpa’s watch loses 30 minutes every hour. If at midnight both set their watches to the correct time, what time will it be when the watches next agree? Solution
12:00 p.m. The two watches will agree only if the times they show differ by some multiple of 12 hours (to include a zero difference)— for example, 4 a.m. and 4 p.m. Let x equal the number of hours elapsed since midnight. The number of hours based on the time shown on Grandma’s watch is modeled by 1.5x and on the time shown on Grandpa’s watch is modeled by 0.5x. The difference between the number of hours shown is modeled as 1.5x – 0.5x = x. Set 12n = x and observe that n = 0 → x = 0 models the statement that the watches agree initially (at midnight). The equation n = 1 → x = 12 implies that 12 hours elapse before the watches next agree. Twelve hours from midnight is 12 p.m. The watches will each show 6:00.
From Calendar Problems , October 2008

 Dizzy Die A die whose dots on opposite faces total 7 lies on a table with the 1 facing east, as shown. The die is rotated four times, each time 90º about an edge. Faces in contact with the table will be 1, then 2, then 3, and finally 5. In what direction does the 1 face after these moves?
Solution
West. Given the value of the face that will be in contact with the table surface, the die will be rotated 90° east, then east, then south, and then south about an edge. The two eastern rolls will move the face showing the 1 a total of 180°, such that the 1 now faces west. The southern rolls will have no impact on the direction of the face showing the 1.
From Calendar Problems , October 2008

 How many? How many fourdigit positive integers x are there with the property that x and 3x have even digits only? (One such number is x = 8002, since 3x = 24,006 and x and 3x each have even digits only.) Solution
82. For both x and 3x to have even digits only, x must be in one of the following forms, where a, b, and c represent digits that may take on values from among {0, 2, 8} and n may take on values from among {2, 8}:
In total, there are 82 possibilities for x.
From Calendar Problems , December 2008

 Chickens and Pigs A farmer has some pigs and some chickens. He sent his son and his daughter to count how many of each he has. "I counted seventy heads," said his son. "And I counted two hundred legs," said his daughter. How many pigs and how many chickens does the farmer actually have? Solution
Thirty pigs and forty chickens. Each chicken has 2 legs, and each pig has 4 legs. Since the number of heads is 70, the number of legs must be 140 plus 2 additional legs for each pig. Which account for the remaining 60 legs. Thus, the farmer has thirty pigs and forty chickens.
From May Calendar, Mathematics Teacher, May 2000

 Upright Integers An integer is defined as upright if the sum of its first two digits equals its third digit. For example, 145 is an upright integer since 1 + 4 = 5. How many positive threedigit integers are upright? Solution
45. From the definition, the first and second digits of an upright integer automatically determine the third digit, which is the sum of the first two digits. Consider those upright integers beginning with 1: 101, 112, 123, 134, 145, 156, 167, 178, and 189; there is a total of 9 such numbers. (Note that the second digit may not be 9; otherwise, the last “digit” would be 1 + 9 = 10.) Beginning with 2, the upright integers are 202, 213, 224, 235, 246, 257, 268, and 279; there is a total of 8 such numbers. We may continue this pattern of analysis to show that the numbers of upright integers beginning with a digit of 3, 4, 5, 6, 7, 8, or 9 are 7, 6, 5, 4, 3, 2, and 1, respectively. Therefore, there is a total of 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 threedigit upright integers.
From Calendar Problems , December 2008

 Wire Length In the diagram, the rectangular wire grid contains 15 identical squares. The length of the grid is 10. What is the length of wire needed to construct the grid?
Solution
76. Since the length of the rectangular grid is 10, the side of each square in the grid is 10 ÷ 5 = 2. The height of the grid is therefore 6 (3 squares). There are four horizontal wires, each of length 10, and six vertical wires, each of length 6, for a total length of wire of 4 • 10 + 6 • 6 = 40 + 36 = 76.
From Calendar Problems , December 2008

 Tires tires tires A threewheeled vehicle travels 100 km. Two spare tires are available. Each of the five tires is used for the same distance during the trip. For how many kilometers is each tire used? Solution
60. Since only three of the five tires are in use at any time, the total distance traveled by all the tires is 3(100) = 300 km. However, each of the five tires is used for the same distance during the trip. Thus, each tire is used for 300 ÷ 5 = 60 km.
From Calendar Problems , December 2008

 Counterfeit Coins III Five counterfeit coins are mixed with nine authentic coins. If two coins are drawn at random, find the probability that one coin is authentic and one is counterfeit. Solution
45/91. If the counterfeit coin is chosen first, the probability is 5/14 • 9/13. If the authentic coin is chosen first, the probability is 9/14 • 5/13. The total probability is 5/14 • 9/13 + 9/14 • 5/13 = 45/91.
From Calendar Problems , August 2008

 Counterfeit Coins II Of 80 coins of the same denomination, one is counterfeit and lighter than the others. Identify the counterfeit coin by using a pan balance in no more than four weighings. Solution
First, make three groups of coins consisting of 27, 27, and 26 coins. Place one group of 27 coins on each balance pan. If the scale does not balance, the counterfeit coin is among the coins in the lighter group. If the scale balances, the counterfeit coin is among the group of 26 coins. Therefore, the problem is reduced to the following: Given 27 coins, find a lighter counterfeit coin by using three weight trials. (If the counterfeit coin is among the 26 coins, add a genuine coin to make 27 coins.) For the second weighing, divide the 27 coins into three groups of 9 each and identify the group of 9 containing the lighter counterfeit coin. For the third weighing, divide the group of 9 containing the counterfeit coin into three groups of 3 coins each. Identify the group of 3 containing the counterfeit coin. Finally, weighing 2 of the 3 coins will identify the counterfeit coin.
From Calendar Problems , September 2008

 Counterfeit Coin You have twelve coins, one of which is counterfeit and weighs less than the legal coins. How can you use a simple balance three times to determine which coin is counterfeit? Solution
One solution is to begin by labeling the coins with the letters from A to L. First, we "weigh" coins A, B, C, and D on one side of the balance and coins E, F, G, and H on the other side. If both sides weigh the same, the fake coin is either I, J, K, or L. If one side is lighter, it includes the counterfeit coin. Thus, after one weighing, we know that the counterfeit coin is one of four. Putting two of the coins from the lightest set on one side of the scale and two on the other, we learn which pair is lighter than the other, so we know which pair contains the counterfeit coin. We use the third weighing to compare these two coins—the lighter one is the counterfeit coin.
From October Calendar, Mathematics Teacher, October 2001

 Guessing Andy is taking a multiplechoice test consisting of 10 questions. All the items have four answer choices but only one correct answer. Unfortunately, Andy did not study for the test and plans on guessing the answer for each item. What is the probability that Andy will guess the answer for every item correctly? Solution
≈ 0.0001%;
(1/4)^{10} = 1/1,048,576
or about 0.0001%, or about one in a million.
From March Calendar, Mathematics Teacher, March 2008

 Who Pays for Dinner?Daniel is having dinner with a friend. He
buys five dishes and his friend buys three
dishes. At the last minute another friend
comes to eat with them and pays $4 for his
share of the meal. If all the dishes have the
same value, how should the money be divided
between Daniel and his first friend? Solution Since the second friend paid $4, the total cost of the meal must be $4 • 3 = $12.
Eight dishes were eaten, so each one costs
$1.50. Daniel brought 5 • $1.50 = $7.50
worth of food, and since his share was $4,
he should receive $3.50.
The first friend brought 3 • $1.50 = $4.50 worth of food and should receive 50 cents.
From May Calendar, May 2005 MT.
From 100 Numerical Games by Pierre Berloquin (Barnes & Noble, 1994).

 House PricesThe average house price in Boomtown rose 30 percent each year for the last five years. If the average house price is currently $250,000, what was the average house price five years ago? Solution Approximately $67,332. If P was the average house price five years ago, then the current average price is 1.3^{5P or 3.71293P. Thus, 250,000 = 3.71293P, so P=250,000/3.71293, which equals $67,332.27From February Calendar, February 2002} 
 999 Coins Starting with a single pile of 999 coins, a person does the following in a series of steps: In step one, he splits the pile into two nonempty piles. Thereafter, at each step, he chooses a pile with 3 or more coins and splits this pile into two piles. What is the largest number of steps that is possible? Solution
997. The number of steps is one less than the number of piles, and 998 is the largest number of piles, 997 with 1 coin and 1 with two coins.
From January Calendar, January 2007 MT

 Perfect Squares and Cubes...NOT How many numbers from 1 to 1 million, inclusive, are not perfect squares or perfect cubes? Solution
998,910. There are 1000 perfect squares between 1 and 1 million; these are the squares of the first 1000 integers. Similarly, there are 100 perfect cubes—the cubes of the numbers from 1 to 100. Subtract the squares and the cubes from 1 million to get 998,900. However, every number that is a perfect sixth power has been subtracted twice (the largest of these is 10^{6} = 1,000,000). Adding these back in gives 998,910.
From Calendar Problems , September 2008

 External Tangent Two circles with radii of 8 cm and 3 cm, respectively, have a common external tangent that determines a segment AB of length 12 cm between the points of tangency. What is the distance between the centers of the circles?
Solution
13 cm. AB is a common external tangent to circles R and T. RA and TB are perpendicular to AB at the points of tangency. Construct a line perpendicular to RA from point T, locating point C. ΔRCT is a right triangle having legs TC (12 cm) and RC (8 – 3 = 5 cm). Therefore, (RT)² = 5² + 12² = 169 and RT = 13 cm.
From March Calendar, Mathematics Teacher, November 2005

 At Random A threedigit number is drawn at random from all possible 3digit numbers formed by the digits 1, 2, 3, 4, and 6. What is the probability that the number drawn is an even number less than 500 containing no digit more than once? Solution
.24. There are 5 × 5 × 5, or 125, different 3digit numbers that can be formed from the 5 given digits. The event requires that the last digit be even and the value of the threedigit number be less than 500. If a 6 is in the units position there are 4 × 3 × 1 differentdigit numerals of this type. If a 2 or a 4 is in the units position, there are 3 × 3 × 2 differentdigit numerals of this type. The number of events therefore is 4 × 3 × 1 + 3 × 3 × 2 = 30. The probability of the event is 30/125 = 6/25 = .24.
From March Calendar, Mathematics Teacher, November 2005

 Divisible by 13 Find the smallest positive integer that is divisible by 13 and that when divided by any of the integers from 2 to 12 (inclusive) leaves a remainder of 1. Solution
83161. The number sought must be of the form K × M + 1 = 13N, where K and N are positive integers and M is the least common multiple of the set of integers 2 through 12. Since the LCM (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) is 27720, the expression becomes K × 27720 + 1 = 13N, and the smallest positive value of K that satisfies the given conditions is 3. For K = 3, 13N = 83161. Therefore, the number is 83161.
From March Calendar, Mathematics Teacher, November 2005

 A Quarter Pounder A quarterpound hamburger contains approximately 80 calories per ounce of meat, an average french fry contains about 14 calories, a cola contains about 10 calories per ounce, and a bun contains 200 calories. Suppose you have a quarterpound hamburger with a bun and six ounces of cola. How many french fries can you eat and still keep your meal below 800 calories? Solution
15. We have 80h + 14f + 10c + 200 < 800. Using the information in the problem we can write 80(4) + 14f + 10(6) + 200 < 800. This implies that 14f < 220 and f < 16. Therefore, you can eat 15 french fries.
From March Calendar, Mathematics Teacher, November 2005

 Sum of Cubes If a + b = 2 and a² + b² = 5, then what is the value of a³ + b³? Solution
11. Squaring a + b = 2, 2ab = (a + b)² – (a² + b²) = 4 – 5 = –1, so that ab = –1/2. Thus a³ + b³ = (a + b)(a² – ab + b²) = 2⋅(5 + 0.5) = 11.
From October Calendar, Mathematics Teacher, October 2005

 Function Suppose that f(x) = ax + b, where a and b are real numbers. Given that f(f(f(x))) = 8x + 21, what is the value of a + b? Solution
5. Since f(x) = ax + b, we have f(f(x)) = a(ax + b) + b = a²x + b(a + 1) and f(f(f(x))) = a(a²x + b(a + 1)) + b = a³x + b(a² + a + 1). Since this must equal 8x + 21, we deduce that a = 2 and b = 21/7 = 3. Hence, a + b = 5.
From October Calendar, Mathematics Teacher, October 2005

 AD and BD What is the sum of the distances AD and BD in the figure shown?
Solution
27. By the Pythagorean theorem, BD = 10. Let E be a point on AB with ∠DEA = 90°. Then DE = 8, and AE = 21 – 6 = 15. By the Pythagorean theorem, AD = 17. Hence, 10 + 17 = 27.
From October Calendar, Mathematics Teacher, October 2005

 Dog Show Of the animals entered in a dog show, the number of poodles is at least onefifth of the number of beagles and at most onesixth the number of collies. The number of dogs that are poodles or beagles is at least 23. Find the minimum number of collies entered in the show. Solution
c ≥ 24. Let p, b, and c represent poodles, beagles, and collies, respectively. Then
(1/6)c ≥ p ≥ (1/5)b
and
p + b ≥ 23.
Since p ≥ (1/5)b, 5p ≥ b, and 6p ≥ b + p ≥ 23,
6p ≥ 23 and p ≥ 23/6.
But p must be a natural number, so p ≥ 4. Since
(1/6)c ≥ p ≥ 4
we have (1/6)c ≥ 4 and c ≥ 24.
From October Calendar, Mathematics Teacher, November 2005

 Local Taxes Alicia earns $20 per hour, of which 1.45 percent is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes? Solution
$0.29. Since $20 is 2000 cents, she pays (0.0145)(2000) = 29 cents per hour in local taxes.
From March Calendar, Mathematics Teacher, August 2005

 Less than 2005° The sum of the interior angles of a polygon is less than 2005°. What is the largest possible number of sides of the polygon? Solution
13. The sum of the interior angles of a polygon = 180°(n – 2) < 2005° → 180°n < 2365° → n < 13.139. Since n must be an integer, the largest value for n is 13.
From March Calendar, Mathematics Teacher, August 2005

 One or Ten? Given that they are made of the same material, which is heavier: a ball with a radius of 10 inches or 10 balls each with a radius of 1 inch? Solution
The ball with a radius of 10 inches is heavier. Since the volume of a sphere is (4/3)πr^{3}, one ball with a radius of 10 inches would have a volume of (4/3)π⋅10^{3} = (4000/3)π in^{3}. Thus 10 balls of radius 1 inch would have an accumulated volume of only
10⋅[(4/3)π⋅1^{3}] = (40/3)π in^{3}.
From March Calendar, Mathematics Teacher, August 2005

 How Many Pairs? For how many pairs of positive integers (x, y) is x + 2y = 100? Solution
49 pairs. The value of x = 100 – 2y is a positive integer when 100 – 2y > 0. This is true for each positive integer y with 1 ≤ y ≤ 49. Thus there are 49 pairs.
From March Calendar, Mathematics Teacher, August 2005

 Tamika > Carlos?Tamika selects two different numbers at random from the set {8, 9, 10} and adds them. Carlos takes two different numbers at random from the set {3, 5, 6} and multiplies them. What is the probability that Tamika's result is greater than Carlos's result?
(A) 4/9 (B) 5/9 (C) 1/2 (D) 1/3 (E) 2/3 Solution (A). Tamika can get the numbers 8 + 9 = 17, 8 + 10 = 18, or 9 + 10 = 19. Carlos can get 3 × 5 = 15, 3 × 6 = 18, or 5 × 6 = 30. The possible ways to pair these numbers are (17, 15), (17, 18), (17, 30), (18, 15), (18, 18), (18, 30), (19, 15), (19, 18), and (19, 30). Four of these nine pairs show Tamika with a higher result, so the probability is 4/9.
From September Calendar, Mathematics Teacher, September 1999

