Difference between revisions of "1993 AJHSME Problems/Problem 20"
Mrdavid445 (talk | contribs) (Created page with "==Problem== When <math>10^{93}-93</math> is expressed as a single whole number, the sum of the digits is <math>\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qqua...") |
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<math>\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833</math> | <math>\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833</math> | ||
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| + | ==Solution== | ||
| + | |||
| + | <cmath>\begin{align*} | ||
| + | 10^2-93&=7\\ | ||
| + | 10^3-93&=907\\ | ||
| + | 10^4-93&=9907\\ | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | This can be generalized into <math>10^n-93</math> is equal is <math>n-2</math> nines followed by the digits <math>07</math>. Then <math>10^{93}-93</math> is equal to <math>91</math> nines followed by <math>07</math>. The sum of the digits is equal to <math>9(91)+7=819+7=\boxed{\text{(D)}\ 826}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1993|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:11, 5 July 2013
Problem
When 
is expressed as a single whole number, the sum of the digits is
Solution
This can be generalized into 
is equal is 
nines followed by the digits 
. Then is equal to 
nines followed by . The sum of the digits is equal to 
.
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
