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}{ | + | <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>. | + | 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:46, 10 March 2015
Problem 3
If and are digits for which
then
Solution
Working from right to left, we see that . Clearly if is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from . Borrow from the digit , and you get , giving .
Since was borrowed from , we have from the tens column . Again for single digit integers this will not work. Again, borrow from , giving . Solving for :
Finally, since was borrowed from the hundreds column, we have , giving .
As a check, the problem is , which is a true sentence.
The desired quantity is , and the answer is .
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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