Difference between revisions of "1969 AHSME Problems/Problem 20"
(Created page with "== Problem == Let <math>P</math> equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in <math>P</math> is: <math>\text{(A) } 36\quad \t...") |
Serpent 121 (talk | contribs) (→Solution 2) |
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\text{(E) } 32</math> | \text{(E) } 32</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <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. | ||
| + | |||
| + | Note: this applies for numbers <math>33--->99</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. | ||
| + | |||
| + | -serpent_121 | ||
== See also == | == See also == | ||
{{AHSME 35p box|year=1969|num-b=19|num-a=21}} | {{AHSME 35p box|year=1969|num-b=19|num-a=21}} | ||
| − | [[Category: | + | [[Category: Introductory Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 08:51, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
