1971 AHSME Problems/Problem 12

Problem

For each integer $N>1$ , there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by $N.$ If $69,90$ , and $125$ are congruent in one such system, then in that same system, $81$ is congruent to

$\textbf{(A) }3\qquad \textbf{(B) }4\qquad \textbf{(C) }5\qquad \textbf{(D) }7\qquad \textbf{(E) }8$

Solution

The "mathematical system" being alluded to is modulo $N$ , so we shall be using such notation for convenience. From the problem, we know that $69 \equiv 90 \equiv 125 \mod N$ , so the difference between each of these numbers must be a multiple of $N$ . $90-69=21=3\cdot7$ , and $125-90=35=5\cdot7$ . The only common factors between these two numbers are $1$ and $7$ , but we know from the problem that $N\neq1$ . Thus, $N=7$ . Now, $81 \equiv 77+4 \equiv 4 \mod 7$ , so our answer is $\boxed{\textbf{(B) }4}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1971 AHSME Problems/Problem 12

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