Difference between revisions of "1994 AJHSME Problems/Problem 25"
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Notice that: | Notice that: | ||
− | <math>9 | + | <math>9 \cdot 4 = 36</math> and <math>3+6 = 9 = 9 \cdot 1</math> |
− | <math>99 | + | <math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math> |
− | So the sum of the digits of x 9s times x 4s is simply <math>x | + | 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>). |
− | Therefore the answer is <math>94 | + | Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math> |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=24|after=Last <br /> Problem}} | {{AJHSME box|year=1994|num-b=24|after=Last <br /> Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:37, 25 October 2020
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 ).
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.