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Difference between revisions of "1994 AJHSME Problems/Problem 8"

(Created page with "==Problem== For how many three-digit whole numbers does the sum of the digits equal <math>25</math>? <math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{...")
 
 
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math>
 
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math>
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==Solution==
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Because <math>8+8+8=24</math>, it follows that one of the digits must be a <math>9</math>. The other two digits them have a sum of <math>25-9=16</math>. The groups of digits that produce a sum of <math>25</math> are <math>799, 889</math> and can be arranged as follows
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<cmath>799,979,997,889,898,988</cmath>
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The number of configurations is <math>\boxed{\text{(C)}\ 6}</math>.
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==See Also==
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{{AJHSME box|year=1994|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 00:13, 5 July 2013

Problem

For how many three-digit whole numbers does the sum of the digits equal $25$?

$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution

Because $8+8+8=24$, it follows that one of the digits must be a $9$. The other two digits them have a sum of $25-9=16$. The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows

\[799,979,997,889,898,988\]

The number of configurations is $\boxed{\text{(C)}\ 6}$.

See Also

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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