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Difference between revisions of "1997 AJHSME Problems/Problem 5"

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Writing out all the two digit multiples of <math>7</math>, you get <math>14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,</math> and <math>91</math>.  Again you find <math>28</math> and <math>91</math> have a digital sum of <math>10</math>, giving answer <math>\boxed{\textbf{(A)}}</math>.
 
Writing out all the two digit multiples of <math>7</math>, you get <math>14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,</math> and <math>91</math>.  Again you find <math>28</math> and <math>91</math> have a digital sum of <math>10</math>, giving answer <math>\boxed{\textbf{(A)}}</math>.
  
−
You may notice that adding <math>7</math> either increases the digital sum by <math>7</math>, or decreases it by <math>2</math>, depending on whether there is carrying or not.
+
You may notice that adding <math>7</math> either increases the digital sum by <math>7</math>, or decreases it by <math>2</math>, depending on whether there is carrying or not.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:08, 9 July 2020

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 (brute force)

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.

See also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AJHSME/AMC 8 Problems and Solutions

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