Difference between revisions of "1987 AJHSME Problems/Problem 8"
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9 & 8 & 7 & 6 \\ | 9 & 8 & 7 & 6 \\ | ||
& A & 3 & 2 \\ | & A & 3 & 2 \\ | ||
| − | + | & B & 1 \\ \hline | |
\end{tabular}</cmath> | \end{tabular}</cmath> | ||
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The minimum possible value of this sum is when <math>A=B=1</math>, which is <cmath>9876+132+11=10019</cmath> | The minimum possible value of this sum is when <math>A=B=1</math>, which is <cmath>9876+132+11=10019</cmath> | ||
| − | The largest possible value of the sum is when <math>A=B=9</math>, making the sum <cmath>9876+ | + | The largest possible value of the sum is when <math>A=B=9</math>, making the sum <cmath>9876+932+91=10899</cmath> |
| − | Since all the possible sums are between <math>10019</math> and <math> | + | Since all the possible sums are between <math>10019</math> and <math>10899</math>, they must have <math>5</math> digits. |
<math>\boxed{\text{B}}</math> | <math>\boxed{\text{B}}</math> | ||
| Line 26: | Line 26: | ||
{{AJHSME box|year=1987|num-b=7|num-a=9}} | {{AJHSME box|year=1987|num-b=7|num-a=9}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:47, 20 September 2018
Problem
If 
and 
are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is
![\[\begin{tabular}[t]{cccc} 9 & 8 & 7 & 6 \\ & A & 3 & 2 \\ & B & 1 \\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/c/3/e/c3ec4b286485f6102f4f920630fb4e8576991a0d.png)
Solution
The minimum possible value of this sum is when 
, which is ![\[9876+132+11=10019\]](http://latex.artofproblemsolving.com/b/8/d/b8d9c5ff0d72788fd3bafee806fea3330f82d1b0.png)
The largest possible value of the sum is when 
, making the sum ![\[9876+932+91=10899\]](http://latex.artofproblemsolving.com/2/b/f/2bfdb518d8f419c48d2fd270e57ab33b3516fb3e.png)
Since all the possible sums are between 
and 
, they must have 
digits.
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
