1969 AMC 12 Problem 20

Attempt Problem 20 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

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20.

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:

36

35

34

33

32

Bound the factors between

3.6\cdot10^{18}

and

3.7\cdot10^{18},

and between

3.4\cdot10^{14}

and

3.5\cdot10^{14}

Show the product lies between

10^{33}

and

10^{34}

The factors satisfy

\begin{aligned} 3.6\cdot10^{18}&\lt A\lt3.7\cdot10^{18},\\ 3.4\cdot10^{14}&\lt B\lt3.5\cdot10^{14}. \end{aligned}

Hence

12.24\cdot10^{32}\lt AB\lt12.95\cdot10^{32},

10^{33}\lt P\lt10^{34}.

Therefore

P

has

34

digits.

Therefore, the correct answer is C.

Problem 20 in other years