Problem of the Week 2025-2026

Problem of the Week #1 (8/27-9/2):

• River Wood Junior High School uses the following master schedule for scheduling each of the students in the school. The first 5 classes listed (including lunch) are mandatory; that is, everyone must take these classes. The last 4 classes are electives and students may choose 2 of these classes. An asterisk (*) indicates what periods the classes are available. Using the information in the master schedule, create a schedule for Stacey. She wants to take Keyboarding and Spanish as her electives. She is also involved in Student Council and they meet during the 1" lunch period. What will Stacey's schedule look like?

Answer:

The solution can be found by looking at which classes are limited to one specific period based on what gets filled up within the schedule. Firstly, since student council meets during Period 4, Stacey must have Period 4 lunch. Then, the only open period for English is Period 2, so her schedule is (2. English, 4. Lunch). Then, the only open period for Keyboarding is Period 6 (2. English, 4. Lunch, 6. Keyboarding). The only remaining period for Science is Period 7 (2. English, 4. Lunch, 6. Keyboarding, 7. Science). The only remaining period for Math is Period 1 (1. Math, 2. English, 4. Lunch, 6. Keyboarding, 7. Science). Both PE and Spanish can be Periods 3 or 5, so there are two final correct schedules.

Final Schedule: 1. Math, 2. English, 3. PE or Spanish, 4. Lunch, 5. PE or Spanish, 6. Keyboarding, 7. Science.

Winners: Jonathan Wu, Dhyan Patel, Jiawei Huang

Problem of the Week #2 (9/3-9/9):

1. A + B + E + G + J + K + R + S + X + Z = 11

2. 2A + 3B - E + G = 35

3. K - R + S = 6

4. 5E + A - B + Z = -17

5. X + J + K + R = -8

6. G + A + S - Z = 17

7. B + E + J + X = 5

8. A + G + R + E = 2

9. K + S + B - A = -3

10. Z + X + E + J = 2

Solve for all variables. Remember to show work!

Answer:

The solution can be found using the elimination and substitution methods for solving algebraic equations. Use these methods to find variables, then substitute the values of those variables back into the equations to solve remaining variables.

Solution: A=6, B=4, E=-4, G=7, J=3, K=6, R=-7, S=5, X=2, Z=1

Winners: Due to the domain change, the problem was lost to history.

Problem of the Week #3 (9/10-9/16):

Angelina and her friends formed a club named the Extremely Cool Club. They wanted to assign a unique 6-digit secret code number to each member of the club. They decided to use the digits 1, 2, 3, 5, 7, and 9 for their numbering system and each of these digits can appear only once in every secret code number (i.e. 123579 is a valid number, but 113355 is not a valid number). What is the maximum number of members who could join the club if everyone is to be assigned a unique secret code number?

Answer:

Take the first digit of the code. There are six possible numbers that could be assigned to that position. Once a number is assigned to that position, it cannot be reused, so there are only five possible numbers to assign to the second position. That means there are 6*5=30 combinations for the first two digits. There are then 4 possible numbers to assign to the third position, making 6*5*4 combinations. This pattern continues through every digit, leaving 6! combinations. 6! = 6*5*4*3*2*1 = 720 combinations.

Solution: There are 720 combinations.

Winners: Brayden Zhang

Problem of the Week #4 (9/17-9/23):

How many squares are on an 8x8 checkerboard?

Note: this is not asking how many tiles there are, since there are obviously 8x8=64 tiles. Squares include 2x2 squares, 5x5 squares, etc., so there are far more than 64 squares.

Answer:

Find a pattern for smaller boards. 1x1 has 1 square. 2x2 has 5 squares (4+1). 3x3 has 14 squares (9 + 4 + 1). Notice how adding an extra row and column of squares increases the number of squares by the square of the current side length (e.g. increasing to a 3x3 adds 3^2=9 squares). Continue this pattern to an 8x8 square to get 8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^1 = 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204 squares.

Solution: There are 204 squares.

Winners: Jiawei Huang, Brayden Zhang, Samuel Strawhacker

Problem of the Week 2024-2025

Problem of the Week #1 (9/4-9/11):

• If we call an integer boring if all of its digits are the same, then how many integers greater than 1 and less than 10000 are both boring and prime? 

Answer:

All single-digit positive integers are boring and the single-digit primes, (2,3,5,7) are both prime and boring. All boring two-digit positive integers are (11,22,33,44,55,66,77,88,99) and of these, only 11 is prime. The boring three-digit positive integers are (111,222,….,999) and none of these are prime, since 111 = (3)(17). Similarly, since 1111 = (11)(101) and thus there are no 4-digit positive boring integers that are prime. Hence the only boring positive prime integers between 1 and 10000 are (2,3,5,7,11). So the answer is 5.

Winners: Ayan Abrar, Beau Jud, Nathan Moore

Problem of the Week #2 (9/11-9/18):

• A derangement of the digits 123456 is a rearrangement of the digits so that digit n does not appear in position n. Thus 654321 and 246513 are derangements but 653124 is not because the digit 3 appears in position 3. How many derangements of 123456 have a 3 in position 1 and a 1 in position 6?

Answer:

The total number of rearrangements with 3 in position 1 and 1 in position 6 is 4! = 24. 

A 2 is in the position 2 for 3! rearrangements. Similarly, 4 and 5 are in the correct positions for 3! of the arrangements. The digits 1, 3, and 6 cannot be in the correct position. 

If we take the total to be 4! − 3 · 3! = 6 we are mistaken as we have eliminated values with multiple n in position n multiple times. 

The pair (2, 4) are in positions 2 and 4, respectively for 2! arrangements. The same number of arrangements applies to the pairs (2, 5) and (4, 5). 

The trio (2, 4, 5) are in positions 2,4, and 5 respectively for the 1! arrangement. 

Thus there are 4! − 3 · 3! + 3 · 2! − 1! = 11 such derangements. 

An alternate solution is to carefully list all permissible derangements: 

342561 345261 345621 346521 352641 354261 354621 356241 362541 364521 365241 

Winners: Adam Koltuniuk, Aidan Weiss, Jonathan Wu

Problem of the Week #3 (9/18-9/25):

• Write (2 + sqrt(3))^2017 as A + Bsqrt(3) where A and B are integers. What is the remainder when A is divided by 5?

Answer:

Write (2+sqrt(3))^n = asubn + bsubn(sqrt(3)). Then asub0 = 1, bsub0 = 0, asub1 = 2, bsub1 = 1. We see that (2+sqrt(3))^2 = 7 + 4sqrt(3) and (2+sqrt(3))(asubn + bsubn(sqrt(3)) = (2asubn + 3bsubn) + (asubn + 2bsubn)sqrt(3). 

We build a table of the first few values of asubn, bsubn

n         0 1  2  3   4    5     6

asubn 1 2 7 26 97 362 1351

bsubn 0 1 4 15 56 209 780

Observe that (2 + sqrt(3))^3 * (asubn + bsubn(sqrt(3)) = (26asubn + 45bsubn) + (15asubn + 26bsubn)sqrt(3). There asub(n+3) = 26asubn + 45bsubn and asub(n-3) - asubn = 5(5asubn + 9bsubn) is a multiple of five. Therefore asub(n+3) and asubn leave the same remainder upon division by fire. Thus asub2017 has the same remainder as a sub1. This is 2.

Winners: Tre Parks, Adam Koltuniuk

Problem of the Week #4 (9/25-10/2):

• What is the number of positive integers less or equal to 900 that have exactly one prime factor in common with 900? For example, 28 is such a number as it shares the prime factor 2 with 900, but they have no other prime factors in common. The number 12 is not such a number because it has multiple prime factors in common with 900; the prime factors 2 and 3.

Answer:

Because 900 = 2^2 · 3^2 · 5^2 we are looking for numbers that share exactly one of the factors 2, 3, 5 with 900. 

We add the positive integers less than 900 that are divisible by 2 or 3 or 5. We must then twice subtract integers that are divisible by two of these (2, 3), or (2, 5), or (3, 5). 

The integers divisible by 2, 3, and 5 must be added back three times. So the number is 900/(1/2 + 1/3 + 1/5 − 2(1/6 − 1/10 − 1/15)  + 3 · 1/30)  = 450 + 300 + 180 − 2(150 + 90 + 60) + 3 · 30 = 930 − 600 + 90 = 420

Winner: Adam Koltuniuk

Problem of the Week #6 (10/9-10/23):

• A Right circular cylinder with a radius of 2 feet and height of 8 feet is partially filled with oil. When it is placed horizontally on a level surface, as shown in the figure, the top of the oil is 1 foot above this surface. If the cylinder is then placed in a vertical position, with a circular base resting on a level surface, then what is the height (in feet) of the oil above the base? 

Answer:

When the cylinder is horizontal, the volume of the oil is 8 [sector ABCD], where [R] denotes the area of region R. When the cylinder is vertical, the volume of oil is pi*(2^2)h, where h is the cylinder height.

BD = 1, OB =2 thus OD = 1. Since AO = 2 then <OAD = 30 degrees and <AOC = 120 degrees. Hence [sector AOCB] = pi(2^2)(120/360) = 4pi/3. 

[segment ABCD] = [sector AOCB] - [triangle AOC] = 4pi/3 - sqrt(3). Equating the two oil volumes gives 8(4pi/3 - sqrt(3)) = 4pih and thus h = 8/3 - 2sqrt(3) / pi = 1.564.

Winners: Adam Koltuniuk, Sarthak Shinde

Problem of the Week #7 (10/23-10/30):

• The difference of two positive numbers is 4 and the product of the two numbers is 19. The sum of the two numbers is

Answer:

Let the numbers be x and y, with x > y. Then xy = 19 and x - y = 4. Solving gives y^2 + 4y - 19 = 0 and y = -2 +- sqrt(4 + 19). Since y > 0, then we use the plus branch and the corresponding x is x = 2 + sqrt(4+19). Thus x + y = 2 sqrt(23).

Winners: Adam Koltuniuk, Sarthak Shinde, Rohen Maniandan

Problem of the Week #8 (10/30-11/6):

• A spinner has the numbers 1,2,3,4 which are formed by wedges of angles x◦ , 2x◦ , 3x◦ , and 4x◦ , respectively. The probability that any of the numbers is spun is proportional to the angle of the corresponding wedge. The spinner is spun four times. What is the probability that the sum of the numbers is a perfect square? 

Answer: 

The probability that n is spun is n/10.

The possible squares reached in four spins are 4, 9, 16.

The sum of 4 may be attained by 1, 1, 1, 1. This has probability (.1)^4 = .0001. 

The sum of 16 may be attained by spinning 4, 4, 4, 4. This has probability (.4)^4 = .0256. 

The sum of 9 may be attained by spinning 4, 3, 1,1 in any order: 4, 2, 2, 1 in any order; 3, 3, 2, 1 in any order; or 3, 2, 2, 2 in any order.

There are (4!)/(2!) = 12 way to spin 4, 3, 1, 1. Each has probability .4 * .3 * .1^2 - .01012. The probability of spinning 4, 3, 1, 1 in some order is .0144. 

There are 4!/2! = 12 ways to spin 4,2,2,1. Each has probability 0.4 · 0.2^2 · 0.1 = 0.0016. The probability of spinning 4,2,2,1 in some order is 0.0192. 

There are 4!/2! = 12 ways to spin 3,3,2,1. Each has probability 0.3 2 · 0.2 · 0.1 = 0.0018. The probability of spinning 3,3,2,1 in any order is 0.0216. 

There are 4!/3! = 4 ways to spin 3,2,2,2. Each has probability 0.3 · 0.2 3 = 0.0024. The probability of spinning 3,2,2,2 in any order is 0.0096 .

The probability of spinning a total of 9 is 0.0144 + 0.0192 + 0.0216 + 0.0096 = 0.0648. 

The probability of spinning a perfect square is 0.0001 + 0.0256 + 0.0648 = 0.0905 = 181/2000 

Winners: Adam Koltuniuk, Sarthak Shinde

Problem of the Week #9 (11/6-11/13):

• In how many ways may the letters RHAAEEIIOOUU be arranged in a line so that no two consecutive letters are the same? The arrangements AEIOURHAEIOU and RAEIOUHAEIUO are two such arrangements. 

Answer: 

The number of arrangements with AA and EE is 10!/(2^3). There are (5 2) ways to choose two of the letters to have next to each other.

The total number of arrangements in which no consecutive letters are the same is (12!)/((2!)^5) - (5 1) ((11!)/(2^4)) + (5 2)(10!/(2^3) - (5 3) (9!/(2^2) + (5 4)(8!/2) - (5 5)7!

 = 7!/(2^5) (12 * 11 * 10 * 9 * 8 - 5 * 2 * 11 * 10 * 9 * 8 + 10 * 4 * 10 *9 * 8 - 10 * 8 * 9 + 5 * 16 * 8 - 32) 

= 7! (3 * 11 * 10 * 9 - 5 * 11 * 5 * 9 + 10 * 10 * 9 - 10 * 2 * 9 + 5 *  4 -1) 

= 7! (99 * [30 - 25] + 90 * [10-2] + 20 -1) = 7! (495 + 720 + 19) = 5040 *1234 = 6219360.

Winner: Rohen Maniandan

Problem of the Week #10 (11/13-11/20):

• A number of objects is represented by 43 in base b and by 38 in base B. If B* is the smallest value of B such that these conditions can be satisfied and b* the corresponding value of b, then the product of b* and B* is 

Answer: 

If N is the number of objects then N = 3 + 4b = 8 + 3B. Since the units digit in base B is 8 then B > 8. If B = 9 then N = 35 and b = 8. Thus (b*)(B*) = 72.

Winners: Adam Koltuniuk, Sarthak Shinde

Problem of the Week #11 (11/20-12/4):

• I have twenty 3 cent stamps and twenty 5 cent stamps. Using one or more of these stamps, how many different amounts of postage can I make? 

Answer:

If the postage amount is divided by 3, then the remainder will be 0, 1, or 2. Let A0 be the set with remainder 0, A1 be those with remainder 1 and A2 be those with remainder 2. The set A = (3, 6, ... 60) is in A0. The rest of A0 is found by adding k of the 5C stamps by A, where k is a multiple of 3. If k = 18 we can make (93, 96, ..., 150). The postage values (63, 66, ..., 90) can be made using k = 12. Thus A0 = (3, 6, ..., 150) and 50 postal amounts can be made such that the remainder is 0 when divided by 3. The smallest amount in A1 is 10 and similar calculations show that A1 = (10, 13, ..., 160) giving 51 amounts. Finally A2 = (5, 8, ..., 155) also giving 51 amounts. Hence there are 152 different amounts of postage.

Winners: Adam Koltuniuk, Sarthak Shinde

Problem of the Week #12 (12/4-12/11):

• The circumferences of two circles are consecutive primes p1 and p2, with p1 > p2. p1 and p2 are consecutive primes if there are no primes between p1 and p2. Central angles of 30 degrees subtend arcs of lengths s1 and s2 on the two circles. If s1 + s2 = 3 then p1 - p2 is

Answer:

The arc length subtended by a central angle is the radius times the angle in radians. The circle radii are p1 / 2pi and p2 / 2pi. Thus 3 = s1 + s2 = ((p1+p2)/(2pi))(30pi/180) -> p1+p2 = 36. Hence p1 = 19 and p2 = 17. 

Winners: Adam Koltuniuk, Sarthak Shinde

Problem of the Week #13 (1/15-1/29):

• If a and b are integers greater than 0, then the number of pairs (a,b) that satisfy the equation 12a+b=ab is 

Answer:

First note that a = 1 cannot satisfy the equation. Also note consecutive integers have no common divisor since and divisor of both would be a divisor of their difference and their difference is 1. Solving the given equation gives b = 12a/(a-1) . Since a-1 and a have no common divisor then a 1 must divide 12. Hence a must be 2, 3, 4, 5, 7 or 13 and the corresponding b is found from the equation. A less formal solution would be to try values of a in the equation b = 12a/(a-  1) and note that for a > 13, b is not an integer. 

Winners: Adam Koltuniuk, Sarthak Shinde, Jonathan Wu

Problem of the Week #16 (2/26-3/26):

• If f(x) = px + q and f(f(f(x))) = 8x + 21, and if p and q are real numbers, then p^2 + q^2 is equal to

Answer:

f(f((x)) = p(px + q) = (p^2)x + pq

f(f(f(x)) = (p^3)x + (p^2)q + pq = 8x + 21

p =2 q = 3

Winners: Adam Koltuniuk, Sarthak Shinde, Jonathan Wu

Problem of the Week 2024 January-May

Problem of the Week #1 (1/10-1/17):

• Suppose that f(x) is a one-to-one function with x ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose further that for each value of x, f(x) ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Determine the number of such one-to-one functions for which f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) is a multiple of 3.

Answer:

Suppose that f(x) is a one-to-one function with x ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose further that for each value of x, f(x) ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Determine the number of such one-to-one functions for which f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) is a multiple of 3. 

The total number of one-to-one functions is 10! multiplied by (13 10) = (13·12·11)/6 = 286. The sum of the integers from 1 to 13 is 13(14)/2 = 91. Thus for the sum of the 10 numbers chosen for the range to add up to a multiple of three, the three numbers omitted must add to one more than a multiple of three. This can happen if the three missing numbers modulo 3 are {0, 0, 1}, {0, 2, 2}, or {1, 1, 2}. The range contains 4 numbers congruent to 0 and 2 modulo 3, and one number congruent to 1 modulo 3. 

The number of choices for {0, 0, 1} is (4 2) · (5 1)  = 30. 

The number of choices for {0, 2, 2} is (4 1) · (4 2) = 24. 

The number of choices for {1, 1, 2} is (5 2) · (4 1)  = 40. 

Therefore the number of ways to choose three of the thirteen range elements with a sum congruent to 1 modulo 3 is 30 + 24 + 40 = 94 ways. The ten elements chosen may be permuted in 10! ways, each creating a different one-to-one function. Therefore the number of such one-to-one functions is 94 *10!.

Winners: Christopher Svilenov, Ben Samuelson, Parker Melling

Problem of the Week #2 (1/17-1/24):

• What is the remainder when the sum 1! + 2! + 3! + 4! + 5! + · · ·+ 2015! is divided by 2016? 

Answer:

2016 = 2^5 ·3^2 ·7 so 2016 divides 8!. The sum 8!+9!+· · ·+2015! leaves no remainder upon division by 2016 so the only terms that contribute to the non-zero reminder is 1!+2!+· · · 7! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 = 5913 = 2 · 2016 + 1881. 1881 is the answer.

Winner: Braedon Kohlenberger

Problem of the Week #3 (1/24-1/31):

• The digital sum of a number is found by taking the sum of the digits. If the sum is at least ten then take the sum of these digits. Continue this process until the sum is less than ten. For example, 2 + 0 + 1 + 9 = 12 and 1 + 2 = 3 so the digital sum process applied to 2019 ends at 3. Let p be the smallest prime larger than 2019 for which p + 2 is also prime. Where does the digital sum applied to p(p + 2) end? 

Answer:

We begin by noting that if a number N is of the form N = 9q + r the sum of the digits is of the form 9s + r and the digital sum of N is r. If p is prime then it is not divisible by 3 and so is of the form 3k + 1 or 3k + 2. If p is of the form 3k + 1 then p + 2 = 3k + 3 is divisible by 3 and is not prime. Therefore p is of the form 3k + 2. The expression p(p + 2) = p^2 + 2p + 1 − 1 = (p + 1)^2 − 1. Because p is of the form 3k + 2, p(p + 2) = (3k + 3)^2 − 1 = 9(k + 1)^2 − 1 so the digital sum is 8. Alternatively, search to see that the next twin primes after 2019 are 2027 and 2029. Their product is 4,112,783 and the resulting digital sums are 26 and 8. 

Winners: Nate Oldham, Paarth Agrawal, Allison Svilenov

Problem of the Week #4 (1/31-2/7):

• How many numbers less than 1,000,000,000 (one billion) are divisible by 7 and have the property that the non-zero digits are 5, 1, and 1? 

Answer:

We are looking for combinations of 5 · 10^a + 10^b + 10^c that are divisible by seven, where the exponents are integers from {0, 1, 2, 3, 4, 5, 6, 7, 8}. Consider the remainders when 10^n is divided by seven

 n              0 1 2 3 4 5 6 7 8 

10&^n     1 3 2 6 4 5 1 3 2 

5 · 10^n   5 1 3 2 6 4 5 1 3 

If the 5 is in either the 10^0 or the 10^6 slot, then the 1s must be in slots that add to 2 or 9. This can be 6 + 3 (2 ways:n =3 and 1 or ), 5 + 4 (1 way: n=4 and 5), or 1 + 1 (0 ways: the 5 occupies a slot). Thus there are 2 × 3 = 6 such numbers with 5 in the ones or millions slot. 

If the 5 is in either the 10^1 or the 10^7 slot, then the 1s must be in slots that add to 6. This can be 5 + 1 (2 ways), 4 + 2 (2 ways), or 3 + 3 (0 ways). Thus there are 2 × 4 = 8 such numbers with 5 in the 10^n slot when n = 1 or n = 7.

If the 5 is in either the 10^2 or the 10^8 slot, then the 1s must be in slots that add to 4 or 11. This could be 6 + 5 (1 way), 3 + 1 (4 ways), or 2 + 2 (0 ways). Thus there are 2 × 5 = 10 such numbers with 5 in the 10^n slot when n = 2 or n = 8. 

If the 5 is in the 10^3 slot, then the 1s must be in slots that add to 5 or 12. This could be 6 + 6 (0 ways), 4 + 1 (2 ways), or 3 + 2(4 ways). Thus there are 6 such numbers with 5 in the 10^3 slot. 

If the 5 is in the 10^4 slot, then the 1s must be in slots that add to 8. This could be 6 + 2 (2 ways), 5 + 3 (2 ways), or 4 + 4(0 ways). Thus there are 4 such numbers with 5 in the 10^4 slot. 

If the 5 is in the 10^5 slot, then the 1s must be in slots that add to 3 or 10. This could be 6 + 4 (1 way), 5 + 5 (0 ways), 2 + 1 (4 ways). Thus there are 5 such numbers with 5 in the 10^3 slot. 

Hence, in total, there are 6 + 8 + 10 + 6 + 4 + 5 = 39 such numbers. 

Winners: Paarth Agrawal, Ayanna Bodake, Matthew Iskandar

Problem of the Week #6  (2/14-2/21):

• In a tournament with players seeded 1,2,3,4 the probability that seed a beats seed b is b/(a + b). In the first round of the tournament seed 1 plays seed 4 and seed 2 plays seed 3. The two winners of the first round matches play each other for the championship. To the nearest hundredth what is the probability that seed 1 wins the tournament? 

Answer:

The probability that seed 1 wins in the first round is 4/5. The probability that seed 2 wins in the first round is 3/5 and the probability that seed 1 beats seed 2 is 2/3. The probability that seed 3 wins in the first round is 2/5 and the probability that seed 1 beats seed 3 is 3/4.
Hence the probability that seed 1 beats seed 2 to win the tournament is 4/5 · 3/5 · 2/3  = 8/25 and the probability that seed 1 beats seed 3 to win the tournament is 4/5 · (2/5 · 3/4) = 6/25 . These are mutually exclusive events so the probability that seed 1 wins the tournament is 8/25 + 6/25 = 14/25 = 0.56 

Winners: Nate Oldham, Christopher Svilenov

Problem of the Week #7  (2/21-2/28):

• Memory jog for ‘geometric progression’, ( 2, 4, 8, 16 is an example). If the sides of a right triangle form a geometric progression and the short side has length 1, then the hypotenuse has length...?

Answer:

If k is the common ratio of the progression a, b, c then a^2 + b^2 = c^2 --> a^2 + (ak)^2= (ak^2)^2 --> k^4 - k^2 - 1 = 0. Letting z = k^2 gives z^2 - z - 1 = 0 or z = c = (1 +- sqrt(5))/2.

Winners: Braedon Kohlenberger, Beau Jud, Ben Samuelson

Problem of the Week #8  (2/28-3/6):

• Ten numbered chips are placed in a bowl. Four chips are numbered 1, three are numbered 2, two are numbered 3, and one is numbered 4. A chip is drawn at random from the bowl. If the number on the chip is n then the chip is replaced in the bowl and 5−n chips numbered n are added to the bowl for a second drawing. For example, if the first chip drawn has a 1 then the second drawing will have eight 1s, three 2s, two 3s, and one 4. What is the probability that a chip numbered 3 is drawn on the second draw? 

Answer:

A chip numbered 1 is drawn 4/10 of the time, in which case the probability of drawing a chip with a 3 on the second draw is 2/14 . 

A chip numbered 2 is drawn 3/10 of the time, in which case the probability of drawing a chip with a 3 on the second draw is 2/13 . 

A chip numbered 3 is drawn 2/10 of the time, in which case the probability of drawing a chip with a 3 on the second draw is 4/12 . 

A chip numbered 4 is drawn 1/10 of the time, in which case the probability of drawing a chip with a 3 on the second draw is 2/11 .
The probability that a chip numbered 3 is drawn on the second draw is 4/10 · 2/14 + 3/10 · 2/13 + 2/10 · 4/12 + 1/10 · 2/11 = 565/3003 

Winners: Hasara Jayesekere

Problem of the Week #9  (3/6-3/13):

• A man has walked two-thirds of the distance across a railroad bridge when he sees a train approaching at 54 miles per hour. If R is his uniform speed (in miles per hour) such that he can just manage to escape by running to either end of the bridge, then the sum of the digits in R is 

Answer:

Let 2D be the distance he has walked across the bridge when he first sees the train, D the remaining distance and R his required running rate. Let time equal 0 when he first sees the train. If he runs toward the oncoming train then he will just clear the bridge as the train arrives at time D/R. If he runs the other direction he will exit the bridge at time 2D/R. The train will arrive at this end of the bridge at time D/R plus the time for the train to cross, that is D/ R + 3D/54.
Hence D/R + 3D/54 = 2D/R --> R = 18
Sum of digits of 18 is 9.

Winners: Braedon Kohlenberger

Problem of the Week #11 (3/20-3/27):

• Sara and David were reading the same novel. When Sara asked David what page he was reading, he replied that the product of the page number he was reading and the next page number was 100172. The sum of the digits for the page was David reading was 

Answer: 

The page number will be near the square root of 100172. Note that 300^2 = 90000, 310^2 = 96100, 320^2 = 102400. Thus the sqrt(100172) is between 310 and 320 but closer to 320. Also the product of the (consecutive) units digits on the two pages must have units digit 2. Trying the pairs (6,7) and (8,9) for for the units digits, we find that the page numbers are 316 and 317. 10

Winners: Hasara Jayesekere, Aidan Weiss, Braedon Kohlenberger

Problem of the Week #12 (3/28-4/3):

• A six-sided die has the numbers 1,2,3,4,5,6. Herb rolls the die twice and adds the values. A four sided die has the numbers 1,2,3,4. Zelma rolls this die three times and adds the values. What is the probability that they have the same total? 

Answer:

We compute the probability of each possible sum 

Total 2 3 4 5 6 7 8 9 10 11 12 

Herb 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 

Zelma 0 1/64 3/64 6/64 10/64 12/64 12/64 10/64 6/64 4/64 1/64 

For each sum we add the probability that they both roll the same value 1/(36·64) (2 · 1 + 3 · 3 + 4 · 6 + 5 · 10 + 6 · 12 + 5 · 12 + 4 · 10 + 3 · 6 + 2 · 4 + 1 · 1) = 47/384. 

Winners: Hasara Jayesekere, Braedon Kohlenberger, Nathan Moore

Problem of the Week #13 (4/3-4/10):

• What is the sum of all three-digit numbers whose digits sum to 9?

Answer:

To get a sum of 9 we can have 9 + 0 + 0, k + (9 − k) + 0 for 1 ≤ k ≤ 8, k + k + (9 − 2k) for 1 ≤ k ≤ 4, or j + k + (9 − j − k) with j < k < (9 − j − k). The case 9, 0, 0 The only number is 900 so the sum of these is 900. The case k, 9−k, 0 For k = 1 and k = 8 we have 810+180+801+108 = 990+909 = 1899. Similarly, the sum for other values of k is 990 + 909 = 1899 so the total for this case is 4 · 1899 = 7596. The case k, k, 9 − 2k For k = 1 we have 117 + 171 + 711 = 999. Similarly, for k = 2, 4 the sum is 999. For k = 3, the sum is 333. So the total for this case is 3 · 999 + 333 = 3330. The case j < k < 9−j −k. The only possible distinct triples are 126, 135, and 234. Each of these may be permuted in 6 ways. In the 126 case, we have 126+261+612+162+621+216 = 999 + 999 = 1998. So the total for this case is 3 · 1998 = 5994. The total for all the numbers is 900 + 7596 + 3330 + 5994 = 8496 + 9324 = 17820 

Winners: Hasara Jayesekere, Braedon Kohlenberger