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> | ||
| Line 10: | Line 10: | ||
<math> \mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18 </math> | <math> \mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18 </math> | ||
| − | + | ==Solution== | |
| + | |||
| + | 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>: | ||
| + | |||
| + | <math>10 + a - 1 - 8 = 7</math> | ||
| + | |||
| + | <math>1 + a = 7</math> | ||
| + | |||
| + | <math>a = 6</math> | ||
| + | |||
| + | Finally, since <math>1</math> was borrowed from the hundreds column, we have <math>7 - 1 - 4 = c</math>, giving <math>c = 2</math>. | ||
| + | |||
| + | As a check, the problem is <math>762 - 489 = 273</math>, which is a true sentence. | ||
| + | |||
| + | The desired quantity is <math>a + b + c = 6 + 9 + 2 = 17</math>, and the answer is <math>\boxed{D}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
