Difference between revisions of "2007 AMC 8 Problems/Problem 18"
m (→Solution) |
Pi is 3.14 (talk | contribs) (→Solution) |
||
Line 8: | Line 8: | ||
<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math> | <math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/7an5wU9Q5hk?t=2085 | ||
==Solution== | ==Solution== |
Revision as of 19:18, 27 October 2020
Contents
Problem
The product of the two -digit numbers
and
has thousands digit and units digit . What is the sum of and ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=2085
Solution
We can first make a small example to find out and . So,
The ones digit plus thousands digit is .
Note that the ones and thousands digits are, added together, . (and so on...) So the answer is This is a direct multlipication way.
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.