Difference between revisions of "2014 AMC 10A Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | + | We can list the first few numbers in the form <math>8*(8....8)</math> | |
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| + | <math>8*8 = 64</math> | ||
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| + | <math>8*88 = 704</math> | ||
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| + | <math>8*888 = 7104</math> | ||
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| + | <math>8*8888 = 71104</math> | ||
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| + | <math>8*88888 = 711104</math> | ||
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| + | By now it's clear that the numbers will be in the form <math>7</math>, <math>k-2</math> <math>1</math>s, and <math>04</math>. We want to make the numbers sum to 1000, so <math>7+4+(k-2) = 1000</math>. Solving, we get <math>k = 991</math>, meaning the answer is <math>\fbox{(D)}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 19:18, 11 July 2016
- The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20, so both problems redirect to this page.
Problem
The product 
, where the second factor has 
digits, is an integer whose digits have a sum of 
. What is ?
Solution
We can list the first few numbers in the form 
By now it's clear that the numbers will be in the form 
, 
s, and 
. We want to make the numbers sum to 1000, so 
. Solving, we get 
, meaning the answer is 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AMC 10 Problems and Solutions | ||
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
