Choosing Integers

Number Theory Level 4

There are 15 (not necessarily distinct) integers chosen uniformly at random from the range from 0 to 999, inclusive. Albert then computes the sum of their units digits, while Bob computes the last three digits of their sum. The probability of them getting the same result is m n \frac mn , for relatively prime positive integers m m and n n . Find 10 m + n 10m+n .


The answer is 110.

Sahil P. by Sahil P. , 19, United Kingdom

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1 solution

Patrick Corn
Jan 4, 2018

Suppose the sum of the first 14 numbers' units digits is d . d. Suppose the sum of the first 14 numbers mod 1000 1000 is e . e. Now, for Albert and Bob to get the same result, we are looking for a fifteenth integer n = 10 a + b , 0 ≤ a ≤ 99 , 0 ≤ b ≤ 9 , n = 10a+b, 0 \le a \le 99, 0 \le b \le 9, such that d + b ≡ e + 10 a + b ( m o d 1000 ) . d+b \equiv e + 10a+b \pmod{1000}. This becomes 10 a ≡ d − e ( m o d 1000 ) . 10a \equiv d-e \pmod{1000}. Now d − e d-e is divisible by 10 , 10, so this always has a unique solution a . a. So there are ten possibilities for n = 10 a + b , n = 10a+b, out of one thousand. This means that the probability is 1 100 , \frac1{100}, and the answer is 110 . \fbox{110}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Cud u pls explain why only 10 possibilities for a

SOUVIK PAL - 3 years, 1 month ago

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The number a a is uniquely determined, and then there are ten choices for b . b.

Patrick Corn - 3 years, 1 month ago

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