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 $\boxed{A}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1999 AHSME Problems/Problem 4

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