Difference between revisions of "1994 AJHSME Problems/Problem 25"
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<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math> | <math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Notice that: | Notice that: | ||
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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 | + | 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 values of <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> | ||
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| + | == Solution 2 == | ||
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| + | <cmath>\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6</cmath> | ||
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| + | <cmath>4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{\text{(A)}\ 846}</cmath> | ||
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| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==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}} | ||
Latest revision as of 08:30, 12 January 2024
Contents
Problem
Find the sum of the digits in the answer to
where a string of 
nines is multiplied by a string of fours.
Solution 1
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 values of ~MATHWIZARD10).
Therefore the answer is 
Solution 2
![\[\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6\]](http://latex.artofproblemsolving.com/2/c/4/2c4e35056eeb18ed80d63ee83cbdbb4a5008ae48.png)
![\[4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{\text{(A)}\ 846}\]](http://latex.artofproblemsolving.com/e/6/7/e670aff931f8a870df600bfa69d092ead3454350.png)
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. 
