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Difference between revisions of "1998 AHSME Problems/Problem 3"
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If <math>\texttt{a,b,}</math> and <math>\texttt{c}</math> are digits for which
 
If <math>\texttt{a,b,}</math> and <math>\texttt{c}</math> are digits for which
  
−
<center><math>\begin{tabular}{r}&\ \texttt{7 a 2}\\ &- \texttt{4 8 b} \\  
+
<center><math>\begin{tabular}{rr}&\ \texttt{7 a 2}\\ -& \texttt{4 8 b} \\  
 
\hline  
 
\hline  
 
&\ \texttt{c 7 3} \end{tabular}</math></center>
 
&\ \texttt{c 7 3} \end{tabular}</math></center>
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==Solution==
 
==Solution==
  
−
Working from right to left, we see that <math>2 - b = 3</math>.  Cleaerly if <math>b</math> is a single digit integer, this cannot be possible.  Therefore, there must be some borrowing from <math>a</math>.  Borrow <math>1</math> from the digit <math>a</math>, and you get <math>12 - b = 3</math>, giving <math>b = 9</math>.
+
Working from right to left, we see that <math>2 - b = 3</math>.  Clearly if <math>b</math> is a single digit integer, this cannot be possible.  Therefore, there must be some borrowing from <math>a</math>.  Borrow <math>1</math> from the digit <math>a</math>, and you get <math>12 - b = 3</math>, giving <math>b = 9</math>.
  
 
Since <math>1</math> was borrowed from <math>a</math>, we have from the tens column <math>(a-1) - 8 = 7</math>.  Again for single digit integers this will not work.  Again, borrow <math>1</math> from <math>7</math>, giving <math>10 + (a-1) - 8 = 7</math>.  Solving for <math>a</math>:
 
Since <math>1</math> was borrowed from <math>a</math>, we have from the tens column <math>(a-1) - 8 = 7</math>.  Again for single digit integers this will not work.  Again, borrow <math>1</math> from <math>7</math>, giving <math>10 + (a-1) - 8 = 7</math>.  Solving for <math>a</math>:
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==See Also==
 
==See Also==
 
{{AHSME box|year=1998|num-b=2|num-a=4}}
 
{{AHSME box|year=1998|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 18:46, 10 March 2015

Problem 3

If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which

$\begin{tabular}{rr}&\ \texttt{7 a 2}\\ -& \texttt{4 8 b} \\ \hline &\ \texttt{c 7 3} \end{tabular}$

then $\texttt{a+b+c =}$

$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$

Solution

Working from right to left, we see that $2 - b = 3$. Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$. Borrow $1$ from the digit $a$, and you get $12 - b = 3$, giving $b = 9$.

Since $1$ was borrowed from $a$, we have from the tens column $(a-1) - 8 = 7$. Again for single digit integers this will not work. Again, borrow $1$ from $7$, giving $10 + (a-1) - 8 = 7$. Solving for $a$:

$10 + a - 1 - 8 = 7$

$1 + a = 7$

$a = 6$

Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$, giving $c = 2$.

As a check, the problem is $762 - 489 = 273$, which is a true sentence.

The desired quantity is $a + b + c = 6 + 9 + 2 = 17$, and the answer is $\boxed{D}$.

See Also

1998 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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