Difference between revisions of "2007 AMC 8 Problems/Problem 18"
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| − | answer: b | + | ==Problem== |
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| + | The product of the two <math>99</math>-digit numbers | ||
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| + | <math>303,030,303,...,030,303</math> and <math>505,050,505,...,050,505</math> | ||
| + | |||
| + | has thousands digit <math>A</math> and units digit <math>B</math>. What is the sum of <math>A</math> and <math>B</math>? | ||
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| + | <math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math> | ||
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| + | ==Solution== | ||
| + | We can first make a small example to find out <math>A</math> and <math>B</math>. So, | ||
| + | |||
| + | <math>303\times505=153015 </math> | ||
| + | |||
| + | The ones digit plus thousands digit is <math>5+3=8</math>. | ||
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| + | 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> | ||
| + | This is a direct multiplication way. | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=2085 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/_goaFuScO6M | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2007|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 04:43, 31 December 2022
Problem
The product of the two 
-digit numbers
and 
has thousands digit 
and units digit 
. What is the sum of and ?
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 multiplication way.
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=2085
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
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. 
