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Difference between revisions of "1971 AHSME Problems/Problem 12"

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== Problem ==
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For each integer <math>N>1</math>, there is a mathematical system in which two or more positive integers are defined
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to be congruent if they leave the same non-negative remainder when divided by <math>N.</math> If <math>69,90</math>, and <math>125</math> are
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congruent in one such system, then in that same system, <math>81</math> is congruent to
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<math>\textbf{(A) }3\qquad
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\textbf{(B) }4\qquad
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\textbf{(C) }5\qquad
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\textbf{(D) }7\qquad
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\textbf{(E) }8    </math>
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== Solution ==
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The "mathematical system" being alluded to is [[modulo]] <math>N</math>, so we shall be using such notation for convenience.
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From the problem, we know that <math>69 \equiv 90 \equiv 125 \mod N</math>, so the difference between each of these numbers must be a multiple of <math>N</math>. <math>90-69=21=3\cdot7</math>, and <math>125-90=35=5\cdot7</math>. The only common factors between these two numbers are <math>1</math> and <math>7</math>, but we know from the problem that <math>N\neq1</math>. Thus, <math>N=7</math>. Now, <math>81 \equiv 77+4 \equiv 4 \mod 7</math>, so our answer is <math>\boxed{\textbf{(B) }4}</math>.
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=11|num-a=13}}
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{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 11:15, 1 August 2024

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 $\boxed{\textbf{(B) }4}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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