Difference between revisions of "2014 AMC 10A Problems/Problem 20"
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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> | 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> | ||
| + | |||
| + | Another way to proceed is that we know the difference between the sum of the digits and <math>k</math> is always <math>9</math>, so we just do <math>1000-9=\boxed{\textbf{(D)991}}</math>. | ||
==Solution 2(Educated Guesses if you have no time)== | ==Solution 2(Educated Guesses if you have no time)== | ||
Revision as of 00:06, 30 December 2020
- The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20, so both problems redirect to this page.
Contents
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 
Another way to proceed is that we know the difference between the sum of the digits and is always 
, so we just do 
.
Solution 2(Educated Guesses if you have no time)
We first note that 
and so we assume there are 
8s.
Then we note that it is asking for the second factor, so we subtract (the original 
in the first factor).
Now we have 
The second factor is obviously a multiple of 
.
Listing the first few, we have 
We notice that the 4th answer choice is 1 less than a 992(a multiple of 124.)
Thus we make an educated guess that it is somehow less by 1, so we get . ~mathboy282
Note(Must Read)
We were just lucky; this method is NOT reliable. Please note that this probably will not work for other problems and is just a lucky coincidence.
Video Solution
~savannahsolver
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. 
