1997 AJHSME Problems/Problem 5

Problem

There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is

$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$

Solution 1

Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ . The two numbers on that list that are divisible by $7$ are $28$ and $91$ . Their sum is $28+91=119$ , choice $\boxed{\textbf{(A)}}$ .

Solution 2

Writing out all the two digit multiples of $7$ , you get $14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,$ and $91$ . Again you find $28$ and $91$ have a digital sum of $10$ , giving answer $\boxed{\textbf{(A)}}$ .

You may notice that adding $7$ either increases the digital sum by $7$ , or decreases it by $2$ , depending on whether there is carrying or not.

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1997 AJHSME Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See also