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> | We can list the first few numbers in the form <math>8*(8....8)</math> | ||
| + | |||
| + | (Hard problem to do without the multiplication, but you can see the pattern early on) | ||
<math>8*8 = 64</math> | <math>8*8 = 64</math> | ||
Revision as of 19:27, 9 March 2020
- 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 
(Hard problem to do without the multiplication, but you can see the pattern early on)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
