Difference between revisions of "1998 AHSME Problems/Problem 3"

m (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>

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 (ProblemsAnswer KeyResources)
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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