1978 AHSME Problems/Problem 27

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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