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1978 AHSME Problems/Problem 27: Difference between revisions

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{{AHSME box|year=1978|num-b=26|num-a=28}}
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[[Category: Introductory Number Theory Problems]]

Latest revision as of 19:01, 15 October 2025

Problem 27

There is more than one integer greater than $1$ which, when divided by any integer $k$ such that $2 \le k \le 11$, has a remainder of $1$. What is the difference between the two smallest such integers?

$\textbf{(A) }2310\qquad \textbf{(B) }2311\qquad \textbf{(C) }27,720\qquad \textbf{(D) }27,721\qquad \textbf{(E) }\text{none of these}$


Solution

Let this integer be $n$. We have $n\equiv 1 \mod 2$, $n \equiv 1 \mod 3$, $n \equiv 1 \mod 4$ $\ldots$ $n \equiv 1 \mod 11$. Recall that if \[a \equiv b \mod c\] and \[a \equiv b \mod d\] then \[a \equiv b \mod (lcm (c,d))\] We see that since $n\equiv 1 \mod 2$, $n \equiv 1 \mod 3$, $n \equiv 1 \mod 4$ $\ldots$ $n \equiv 1 \mod 11$. We have \[n \equiv 1 \mod (lcm (2, 3, \ldots , 10, 11))\]

From $2$ to $11$, $8$ contains the largest power of $2$, $9$ contains the largest power of $3$, and $10$ contains the largest power of $5$. Thus, our lcm is equal to \[2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11\] Since $n > 1$, our $2$ smallest values of $n$ are \[n = 1 + (2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11)\] and \[n = 1 + 2(2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11)\] The difference between these values is simply the value of \[(2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11) = 40 \cdot 99 \cdot 7\] \[= 3960 \cdot 7\] \[= 27720, \boxed{C}\]

~JustinLee2017

See Also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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