Difference between revisions of "1994 AJHSME Problems/Problem 25"
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<math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math> | <math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math> | ||
− | So the sum of the digits of <math>x</math> 9s times <math>x</math> 4s is simply <math>x \cdot 9</math> (Try to find the proof that it works for all <math>x</math>). | + | So the sum of the digits of <math>x</math> 9s times <math>x</math> 4s is simply <math>x \cdot 9</math> (Try to find the proof that it works for all <math>x</math> ~MATHWIZARD10). |
Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math> | Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math> |
Latest revision as of 21:33, 18 January 2021
Problem
Find the sum of the digits in the answer to
where a string of nines is multiplied by a string of fours.
Solution
Notice that:
and
and
So the sum of the digits of 9s times 4s is simply (Try to find the proof that it works for all ~MATHWIZARD10).
Therefore the answer is
See Also
1994 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.