# Difference between revisions of "1999 AHSME Problems/Problem 4"

## Problem

Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$. $\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$

## Solution

Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$.

No prime number ends in $4$, since all numbers that end in $4$ are divisible by $2$. Thus, we are only looking for numbers that end in $9$.

Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$.

Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$.

Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$, since all numbers are odd), we get: $29, 49, 89.$

Noting that $49$ is not prime, we have only $29$ and $89$, which give a sum of $118$, so the answer is $\boxed{A}$.

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