1976 AMC 12 Problem 15

Attempt Problem 15 of the 1976 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 1976 AMC 12 solutions, or check the answer key.

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15.

r

is the remainder when each of the numbers

1059,

1417

and

2312

is divided by

d,

where

d

is an integer greater than one, then

d-r

equals

1

15

179

d-15

d-1

A common remainder means

d

divides every pairwise difference

Use

1417-1059=358=2\cdot179

and

2312-1417=895=5\cdot179

The divisor

d

divides both

358=2\cdot179

and

895=5\cdot179.

Their greatest common divisor is

179,

a prime, so

d=179.

Since

1059=5\cdot179+164,

we have

r=164

and

d-r=15.

Therefore, the correct answer is B.

Problem 15 in other years