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

(Created page with "The two numbers 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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The two numbers 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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==Problem==
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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
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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>
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==Solution 1==
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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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==Solution 2==
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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>.
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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. 
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== See also ==
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{{AJHSME box|year=1997|num-b=4|num-a=6}}
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* [[AJHSME]]
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* [[AJHSME Problems and Solutions]]
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* [[Mathematics competition resources]]

Revision as of 13:40, 31 July 2011

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.

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
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