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

Revision as of 18:25, 14 November 2011

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:

$9*4 = 36$ and $3+6 = 9 = 9*1$

$99*44 = 4536$ and $4+5+3+6 = 18 = 9*2$

So the sum of the digits of x 9s times x 4s is simply $x*9$ .

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

@@ 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*4 = 36</math> and <math>3+6 = 9 = 9*1</math>
+<math>99*44 = 4536</math> and <math>4+5+3+6 = 18 = 9*2</math>
+So the sum of the digits of x 9s times x 4s is simply <math>x*9</math>.
+Therefore the answer is <math>94*9 = \boxed{\text{(A)}\ 846}</math>