1. M is 30% of Q, Q is 20% of P, and N is 50% of P. What is M/N?
M =30% x 20% = 6% of P. Therefore, M/N = 6/50 = 12%.
2. The sum of two integers is 13, and the sum of their squares is 85. What is the product of these two integers?
xy = [(x+y)2 – x2 – y2 ] / 2 = (169 -85) = 42
3. In 2013, John’s grandmother was exactly three times his age. The sum of the years in which they were both born is 3,930. How old will John be in 2020?
Let G and J be current ages of Grandma and John respectively. Let BY stand for Birth Year. We know that G = 3J. We also know that BYG + BYJ = 3930. Finally, we know that BYG+ G = BYJ+ J = 2013.
On one hand, (BYG+ G) + (BYJ+ J) = 4026. On the other hand, (BYG+ G) + (BYJ+ J) =
= (BYG + BYJ)+ J + G = 3930 +4J. Therefore, 4J = 4026 – 3930 = 96. J = 24, John’s current age. In 2020, he will be 31.
4. Compute the largest prime divisor of 15! – 13!, where N! = 1·2·3…·N (N-factorial)
15! – 13! = 13! (15*14 – 1) = 13! * 209 = 13! * 11 *19. The largest prime divisors is 19.
5. A 7-digit telephone number d1 d2 d3 d4 d5 d6 d7 is called memorable if the prefix d1 d2 d3 is exactly the same as either d4 d5 d6 or d5 d6 d7 (possibly both). Assume each digit di can be any of the ten decimal digits 0,1,2…9. How many distinct memorable telephone numbers are out there?
There are 1,000 possibilities for d1 d2 d3. The telephone number can become memorable through equality with d4 d5 d6 or d5 d6 d7 or both. If the equality is achieved through d4 d5 d6, then there are still 10 possibilities for d7, and there are a total of 10,000 of such memorable numbers. If the equality is achieved through d5 d6 d7, then there are still 10 possibilities for d4, and there are a total of 10,000 of such memorable numbers. The overlap comes from 10 telephone numbers where all 7 digits are all the same. Total number of possibilities is 10,000 + 10,000 – 10 = 19,990
6. 2564 ·6425 is a square of a positive integer N. What is the sum of decimal digits of N?
If N2 = 2564 ·6425, then N = 564 · 825 =564 · 275 = 564 · 264 · 211 = 211 · 1064 = 2048 · 1064
Besides 64 zeros, the only non-zero digits are 2,4, and 8, as above. Sum of digits is 2+4+8 = 14 .
7. Find the number t and the digit a, such that [3(230+t)]2 = 492,a04.
492,a04 must be divisible by 9, so the sum of digits must be divisible by 9. Therefore, a= 8.
492,804 / 9 = 54, 756 = (230+t)2. As a result, t = 4.
Answer: a=8, t = 4
8. Find the sum of all digits if the integers from 1 to 1,000,000,000 are written down next to each other. That’s all the digits in all the numbers, not all the numbers themselves.
Let’s group all numbers in pairs as follows:
(1 and 999, 999, 998), (2 and 999, 999, 997), etc. The only 2 numbers that don’t have a pair are 999,999,999 and 1,000,000,000. The sum of digits in every pair is 9*9 = 81. So the overall sum of digits, 81* 999,999,998/2 + 1 + 81 = 81 * 500,000,000 + 1 = 40,500,000,001.
9. A cat has decided to take a nap. He dreams that he is encircled by 13 mice: 12 gray and 1 white. He hears his owner saying: “you are to eat each 13th mouse in the clockwise direction. The last mouse you eat must be the white one.” Which mouse should he start from?
Let’s put all mice around the circle and start count with #13, moving clockwise. The mice we will eat will be numbered as follows: 13, 1, 3, 6, 10, 5, 2, 4,9,11,12,7, and 8. We need to set position #8 (relative to the starting point) as white mouse. The ideal starting point relative to the white mouse is, therefore, located 8 spots in the counterclockwise direction or 5 spots in the clockwise direction.
Answer: The cat needs to start with position #5 in the clockwise direction relative to the white mouse.
10. Abraham left city A to do a delivery in city S. Sarah left city S to do a delivery in city A at the same time as Abraham left city A. They met each other 5miles away from city A. Once Abraham and Sarah reached their destination, they immediately turned around and started walking back. They met again 3 miles away from city S. What is the distance between cities A and S?
Let D = total distance between cities.
Prior to the 1st meeting, they jointly covered an entire distance D, while Abraham only covered 5 miles.
Prior to the 2nd meeting, they jointly covered an entire distance 3D (each covered a full distance D and they also shared 1 such such distance). Therefore, the time that passed from the beginning until the 2nd meeting is 3 times larger than the time that passed until the 1st meeting.
This means that Abraham covered a distance of 3*5 = 15 miles prior to the 2nd meeting. Since he was already 3 miles away from city S during 2nd meeting, the distance between cities D = 15 – 3 = 12.
Answer: 12 miles.