2007 AMC 8 Problems/Problem 18: Difference between revisions

Revision as of 18:33, 1 August 2021

The product of the two $99$ -digit numbers

$303,030,303,...,030,303$ and $505,050,505,...,050,505$

has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$ ?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$

We can first make a small example to find out A and B So,

303*505=153015

The ones digit plus thousands digit is 5+3=8

Note that the ones and thousands digits are, added together 8 (and so on...) So the answer is D This is a direct multlipication way.

~savannahsolver

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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 14: / Line 14: @@
 ==Solution==
-We can first make a small example to find out <math>A</math> and <math>B</math>. So,
+We can first make a small example to find out A and B So,
-<math>303\times505=153015 </math>
+*505=153015
-The ones digit plus thousands digit is <math>5+3=8</math>.
+The ones digit plus thousands digit is 5+3=8
-Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math>\boxed{\textbf{(D)}\ 8}</math>
+Note that the ones and thousands digits are, added together 8 (and so on...) So the answer is D
 This is a direct multlipication way.