2019-2020 SMML
A, B, and C are an analyst, a broker, and a carpenter, not necessarily in that order. The broker is married to one of A's siblings. C is not a carpenter. B is unmarried. A is a client of the analyst. Who has what job?
A is the carpenter, B is the analyst, and C is the broker.
A is a client of the analyst and has a sibling who is married to the broker, and so must be the carpenter. B is unmarried and so cannot be the broker, and must therefore be the analyst. C must therefore be the broker.
Write 20215 - 20214 as a base 3 number
111213
20215 = 2*125 + 0*25 + 2 *5 + 1 = 261 and 20214 = 2*64 + 0* 16 + 2*4 + 1 = 137.
261-137 = 124. In base 3, 124 = 1*81 + 1 *27 +1*9 + 2 *3 + 1 = 111213
A certain fast-food restaurant sells chicken nuggets in packs of 4, 6, 9, 10, 20, and 50. What is the largest possible number of chicken nuggets that can NOT be purchased exactly using the above packs?
11
Using 4 and 6 packs, we can eliminate any number of even nuggets starting at 4 (it is either a multiple of 4 or two more than a multiple of 4, which we can get by using a 6 instead of a 4).
Using 9 as a starting point, we can eliminate any number of odd nuggets starting at 13 by using 1 pack of 9 and then packs of 4 and 6 and the logic in the previous statement.
There is no way to purchase exactly 1, 2, 3, 5, 7, or 11 chicken nuggets using the above packs. 11 is the largest possible number.
In what base is 2020b - 2019b = 7
16
2020b - 2019b = 7
(2b3 + 2b) - (2b3 + b + 9) = 7
b - 9 = 7
b = 16
There are three distinct integers a, b, c such that a + b + c = 13, abc = -165, and a2 + b2 + c2 = 155. What are the three integers? (Your answer need not indicate which is a or b or c.)
-3, 5, and 11
The quickest solution starts with finding three integers whose product is -165, and the quickest route to that is the prime factorization of 165, which is 3 * 5 * 11. One or all three of these must be negative if their product is negative, but since their sum is positive, two of them must be positive. And since their sum is 13, the integers must be -3, 5, and 11. Summing their squares validates the solution.
Given positive integers A and B such that 25A + 22B = 43A, find the smallest possible value of 17A + 121B in base 10.
64
Converting the first equation to base 10, we get 2A + 5 + 2B + 2 = 4A + 3. 4 = 2A - 2B. 2 = A - B
The first equation implies that A >/= 6 and B>/= 3, based on the digits present. The second equation implies that A >/= 8 and B >/= 3, based on the digits present. Therefore, the smallest values of A and B based on the last three statements is A = 8 and B = 8.
Converting the second equation to base 10 using these values we get (8 + 7) + (36 + 12 + 1) = 64
There are three pairs of positive integers whose squares differ by 45. What is the sum of the six integers in the three pairs?
69
In other words, (a2 - b2) = 45. This leads to (a + b)(a - b) = 45, which in turn leads to (a - b)(a + b) = 1 x 45 or (a - b)(a + b) = 3 x 15 or (a - b)(a + b) = 5 x 9.
If a - b = 1 and a + b = 45, then a = 23 and b = 22. If a - b = 3 and a + b = 15, then a = 9 and b = 6. If a - b = 5 and a + b = 9, then a = 7 and b = 2. The sum of these six integers is 69.
The fuel tank for a particular engine receives a continuous, steady flow of 20 liters of fuel per hour. The engine consumes a predictable amount of fuel every six hours. Every day, the fuel consumption during these six-hour periods is 100 liters, 175 liters, 133 liters, and 72 liters. The fuel tank must have at least 2 liters of fuel at all times. What is the capacity of the smallest fuel tank that can meet these requirements?
70
The tank receives 120 liters of fuel every six hours, and the engine consumes all 480 liters every day. So if x is the amount of fuel in the tank at the start of each day, the amount of fuel (in liters) in the tank at the end of the first six-hour period is x + 120 - 100 = x + 20. At the end of the second period, the amount remaining is x + 20 + 120 - 175 = x - 35. At the end of the third period, the amount remaining is x - 35 + 120 - 133 = x - 48. Finally, at the end of the fourth period, the amount remaining is x - 48 + 120 - 72 = x.
The smallest of these amounts is x-48, so x has to be at least 50 liters in order to meet the 2-liter-minimum requirement. But 50 liters is too small to hold all the fuel for the first six-hour period, so the tank must have a capacity of 50 + 20 = 70 liters.
Four students are sitting in a row, facing forward, so that the person in Seat #1 can not see any other students, the person in Seat #2 can only see the person in Seat #1, the person in Seat #3 can only see the person in Seat #1 and Seat #2, and the person in Seat #4 can see all of the other students. The students are all wearing a single-colored coat. Use the following information to list the students' names in order, from Seat #1 to Seat #4.
- Students can not see themselves or the coat they are wearing
- Allen can see Wallis
- The person sitting in Seat #2 can NOT see a white coat
- The person wearing a black coat can see the person in Seat #2
- Allen can see the person wearing a black coat
- The person in Seat #1 is NOT wearing a green coat
- Yin is NOT wearing black or red
- Allen is not wearing green
- Jones can NOT see Yin
Jones, Yin, Wallis, Allen