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Difference between revisions of "1995 AHSME Problems/Problem 13"

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==Solution==
 
==Solution==
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If we change 0 only, it would be incorrect. The same for 1, but 2 we could change to 6. If we try all the other digits, we see that it would be impossible to correct the addition. Thus <math>d=2</math> and <math>e=6</math>. <math>d+e=\boxed{8\mathrm{(C)}}</math>
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If we change <math>0</math>, the units column would be incorrect.  
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If we change <math>1</math>, then the leading <math>1</math> in the sum would be incorrect.
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However, looking at the <math>2</math> in the hundred-thousands column, it would be possible to change the <math>2</math> to either a <math>5</math> (no carry) or a <math>6</math> (carry) to create a correct statement.
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Changing the <math>2</math> to a <math>5</math> would give <math>745586 + 859430</math> on top, which equals <math>1605016</math>.  This does not match up to the bottom.
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Changing the <math>2</math> to a <math>6</math> gives <math>746586 + 869430</math> on top, which has a sum of <math>1616016</math>.  This is the number on the bottom if the <math>2</math>s were changed to <math>6</math>s.
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Thus <math>d=2</math> and <math>e=6</math>. so <math>d+e= 8 \boxed{\mathrm{ (C)}}</math>
  
 
==See also==
 
==See also==

Revision as of 00:36, 18 August 2011

Problem

The addition below is incorrect. The display can be made correct by changing one digit $d$, wherever it occurs, to another digit $e$. Find the sum of $d$ and $e$.

$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$

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

Solution

If we change $0$, the units column would be incorrect.

If we change $1$, then the leading $1$ in the sum would be incorrect.

However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.

Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$. This does not match up to the bottom.

Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$. This is the number on the bottom if the $2$s were changed to $6$s.

Thus $d=2$ and $e=6$. so $d+e= 8 \boxed{\mathrm{ (C)}}$

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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