1969 AHSME Problems/Problem 20

Problem

Let $P$ equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in $P$ is:

$\text{(A) } 36\quad \text{(B) } 35\quad \text{(C) } 34\quad \text{(D) } 33\quad \text{(E) } 32$

Solution 1

Through inspection, we see that the two digit number $33^{2}=1089=4$ digits. Notice that any number that has the form $33abcdefg.......$ multiplied by another $33qwertyu.........$ 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 $19$ digits, and the second number has $15$ digits.

Note: this applies for numbers $33--->99$

Hence, the answer is $19+15=34$ digits $\implies \fbox{C}$

Solution 2

We can approximate the product with $10^{18} * 3.6 *10^{14} *3.4$ Now observe that $3.6*3.4>10$ , so we can further simplify the product with $10^{18}*10^{14}*10=10^{33}$ which means the product has $\fbox{34 (C)}$ digits.

-serpent_121

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1969 AHSME Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See also