Difference between revisions of "1969 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | Through inspection, we see that the two digit number <math>33^{2}=1089=4</math> digits. |
| + | Notice that any number that has the form <math>33abcdefg.......</math> multiplied by another <math>33qwertyu.........</math> will have its number of digits equal to the sum of the original numbers' digits. | ||
| + | |||
| + | In this case, we see that the first number has <math>19</math> digits, and the second number has <math>15</math> digits. | ||
| + | |||
| + | Hence, the answer is <math>19+15=34</math> digits <math>\implies \fbox{C}</math> | ||
== See also == | == See also == | ||
Revision as of 15:02, 8 June 2017
Problem
Let 
equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in is:
Solution
Through inspection, we see that the two digit number 
digits.
Notice that any number that has the form 
multiplied by another 
will have its number of digits equal to the sum of the original numbers' digits.
In this case, we see that the first number has 
digits, and the second number has 
digits.
Hence, the answer is 
digits 
See also
| 1969 AHSC (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 • 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. 
