1951 AMC 12 Problem 37

Attempt Problem 37 of the 1951 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1951 AMC 12 solutions, or check the answer key.

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37.

A number which when divided by 1010 leaves a remainder of 9,9, when divided by 99 leaves a remainder of 8,8, by 88 leaves a remainder of 7,7, etc., down to where, when divided by 2,2, it leaves a remainder of 1,1, is:

5959

419419

12591259

25192519

None of these answers

Answer: D
Concepts:least common multiplemodular arithmetic
Difficulty rating: 1580
Small Hint:

Adding 11 to the desired number removes every listed remainder

Big Hint:

Find lcm(2,3,,10)\operatorname{lcm}(2,3,\ldots,10)

Solution:

If the number is N,N, then N+1N+1 is divisible by every integer from 22 through 10.10. Their least common multiple is 233257=2520. 2^3\cdot3^2\cdot5\cdot7=2520. Thus the least positive number fitting all the conditions is N=25201=2519.N=2520-1=2519.

Thus, the correct answer is D.

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