Difference between revisions of "1969 AHSME Problems/Problem 20"

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\text{(E) } 32</math>
 
\text{(E) } 32</math>
  
== Solution ==
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== 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.
  
 
== See also ==
 
== See also ==

Latest revision as of 04:39, 14 February 2020

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.

See also

1969 AHSC (ProblemsAnswer KeyResources)
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

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