Difference between revisions of "1999 AHSME Problems/Problem 6"
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<math>2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}</math>, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be <math>2+5=7</math>, thus making the answer <math>\boxed{\text{D}}</math>. | <math>2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}</math>, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be <math>2+5=7</math>, thus making the answer <math>\boxed{\text{D}}</math>. | ||
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== See also == | == See also == | ||
{{AHSME box|year=1999|num-b=5|num-a=7}} | {{AHSME box|year=1999|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:59, 16 March 2020
Problem
What is the sum of the digits of the decimal form of the product 
?
Solution
, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be 
, thus making the answer 
.
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
