Difference between revisions of "1969 AHSME Problems/Problem 20"
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\text{(E) } 32</math> | \text{(E) } 32</math> | ||
| − | == Solution == | + | == Solution 1 == |
Through inspection, we see that the two digit number <math>33^{2}=1089=4</math> digits. | 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. | 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. | ||
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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> | ||
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| + | == Solution 2 == | ||
| + | We can approximate the product with <math>10^{18} * 3.6 *10^{14} *3.4</math> Now observe that <math>3.6*3.4>10</math>, so we can further simplify the product with <math>10^{18}*10^{14}*10=10^{33}</math> which means the product has <math>\fbox{34 (C)}</math> digits. | ||
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| + | -serpent_121 | ||
== See also == | == See also == | ||
Latest revision as of 07:51, 27 May 2020
Contents
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 1
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 
Solution 2
We can approximate the product with 
Now observe that 
, so we can further simplify the product with 
which means the product has 
digits.
-serpent_121
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. 
