Difference between revisions of "1971 AHSME Problems/Problem 1"
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| − | == Problem | + | == Problem == |
The number of digits in the number <math>N=2^{12}\times 5^8</math> is | The number of digits in the number <math>N=2^{12}\times 5^8</math> is | ||
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== Solution == | == Solution == | ||
| − | <math>N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000</math>, which has <math>2+8=10</math> digits, so the answer is <math> | + | <math>N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000</math>, which has <math>2+8=10</math> digits, so the answer is <math>\boxed{\textbf{(B) }10}</math>. |
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| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 08:25, 1 August 2024
Problem
The number of digits in the number 
is
Solution
, which has 
digits, so the answer is 
.
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
