Difference between revisions of "1969 AHSME Problems/Problem 20"
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In this case, we see that the first number has <math>19</math> digits, and the second number has <math>15</math> digits. | In this case, we see that the first number has <math>19</math> digits, and the second number has <math>15</math> digits. | ||
+ | |||
+ | Note: this applies for numbers <math>33--->99</math> | ||
Hence, the answer is <math>19+15=34</math> digits <math>\implies \fbox{C}</math> | Hence, the answer is <math>19+15=34</math> digits <math>\implies \fbox{C}</math> |
Revision as of 16:03, 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.
Note: this applies for numbers
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.