Difference between revisions of "1969 AHSME Problems/Problem 20"
Serpent 121 (talk | contribs) (→Solution) |
|||
Line 9: | Line 9: | ||
\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. | ||
Line 18: | Line 18: | ||
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> | ||
+ | |||
+ | == 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. | ||
== See also == | == See also == |
Latest revision as of 04:39, 14 February 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.
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.