# 1978 AHSME Problems/Problem 27

## 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

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