Difference between revisions of "1997 AJHSME Problems/Problem 5"
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<math>\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189</math> | <math>\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189</math> | ||
| − | ==Solution 1== | + | ==Solution 1 (brute force)== |
Writing out all two digit numbers that have a digital sum of <math>10</math>, you get <math>19, 28, 37, 46, 55, 64, 73, 82,</math> and <math>91</math>. The two numbers on that list that are divisible by <math>7</math> are <math>28</math> and <math>91</math>. Their sum is <math>28+91=119</math>, choice <math>\boxed{\textbf{(A)}}</math>. | Writing out all two digit numbers that have a digital sum of <math>10</math>, you get <math>19, 28, 37, 46, 55, 64, 73, 82,</math> and <math>91</math>. The two numbers on that list that are divisible by <math>7</math> are <math>28</math> and <math>91</math>. Their sum is <math>28+91=119</math>, choice <math>\boxed{\textbf{(A)}}</math>. | ||
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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 == | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:08, 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
Solution 1 (brute force)
Writing out all two digit numbers that have a digital sum of 
, you get 
and 
. The two numbers on that list that are divisible by 
are 
and . Their sum is 
, choice 
.
Solution 2
Writing out all the two digit multiples of , you get 
and . Again you find and have a digital sum of , giving answer .
You may notice that adding either increases the digital sum by , or decreases it by 
, depending on whether there is carrying or not.
See also
| 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. 
