Difference between revisions of "1987 AJHSME Problems/Problem 8"
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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> |
Revision as of 19:49, 30 May 2017
Problem
If and are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is
Solution
The minimum possible value of this sum is when , which is
The largest possible value of the sum is when , making the sum
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.