Difference between revisions of "1994 AJHSME Problems/Problem 25"

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<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math>
 
<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math>
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==Solution==
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Notice that:
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<math>9 \cdot 4 = 36</math> and <math>3+6 = 9 = 9 \cdot 1</math>
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<math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math>
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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 that it works for all <math>x</math> ~MATHWIZARD10).
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Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math>
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==See Also==
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{{AJHSME box|year=1994|num-b=24|after=Last <br /> Problem}}
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{{MAA Notice}}

Latest revision as of 22:33, 18 January 2021

Problem

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.

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

Solution

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 that it works for all $x$ ~MATHWIZARD10).

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

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last
Problem
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
All AJHSME/AMC 8 Problems and Solutions

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