1978 AHSME Problems/Problem 27

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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1978 AHSME Problems/Problem 27

Solution