1994 AJHSME Problems/Problem 25: Difference between revisions

Revision as of 10:35, 25 October 2020

Find the sum of the digits in the answer to

$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$

where a string of $94$ nines is multiplied by a string of $94$ fours.

$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

Notice that:

$9 \cdot 4 = 36$ and $3+6 = 9 = 9 \cdot 1$

$99 \cdot 44 = 4356$ and $4+5+3+6 = 18 = 9 \cdot 2$

So the sum of the digits of $x$ 9s times $x$ 4s is simply $x \cdot 9$ (Try to find the proof yourself).

Therefore the answer is $94 \cdot 9 = \boxed{\text{(A)}\ 846.}$

1994 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

@@ Line 17: / Line 17: @@
 <math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math>
-So the sum of the digits of <math>x</math> 9s times <math>x</math> 4s is simply <math>x \cdot 9</math>.
+So the sum of the digits of <math>x</math> 9s times <math>x</math> 4s is simply <math>x \cdot 9</math> (Try to find the proof yourself).
 Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math>