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

Revision as of 17:35, 24 April 2021

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

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

1994 AJHSME (Problems • Answer Key • Resources)
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

@@ Line 8: / Line 8: @@
 <math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math>
+==Solution==
+Notice that:
+<math>9 \cdot 4 = 36</math> and <math>3+6 = 9 = 9 \cdot 1</math>
+<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> (Try to find the proof that it works for all values of <math>x</math> ~MATHWIZARD10).
+Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math>
+==See Also==
+{{AJHSME box|year=1994|num-b=24|after=Last <br /> Problem}}
+{{MAA Notice}}