1999 CEMC Gauss (Grade 7) Problems/Problem 2

Problem

The integer $287$ is exactly divisible by

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 6$

Solution

Since $287$ is an odd number, it must not be divisible by any number that is even, eliminating $4$ and $6$ .

For a number to be divisible by $5$ , it must end in $5$ or $0$ . However, $287$ ends in a $7$ , so $287$ is not divisible by $5$ .

The divisibility rule for $3$ states that the sum of the digits of a number must be a multiple of $3$ for the number to be divisible by $3$ . However, the sum of the digits of $287$ is $2 + 8 + 7 = 17$ , which is not a multiple of $3$ .

The only answer choice remaining is $\boxed{\textbf{(D) }7}$ . ~anabel.disher

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1999 CEMC Gauss (Grade 7) Problems/Problem 2

Problem

Solution