Difference between revisions of "2014 AMC 10A Problems/Problem 20"
m (→Solution) |
Captainsnake (talk | contribs) (added formal proof to solution 1 for those who would like a more rigorous version of it.) |
||
Line 24: | Line 24: | ||
Another way to proceed is that we know the difference between the sum of the digits of each product and <math>k</math> is always <math>9</math>, so we just do <math>1000-9=\boxed{\textbf{(D)991}}</math>. | Another way to proceed is that we know the difference between the sum of the digits of each product and <math>k</math> is always <math>9</math>, so we just do <math>1000-9=\boxed{\textbf{(D)991}}</math>. | ||
+ | |||
+ | ===Proof of this solution's validity=== | ||
+ | |||
+ | Since this solution won't fly on a proof-based competition, here's a proof that it's valid: | ||
+ | |||
+ | We will call <math>x_k=8(888\dots8)</math> with exactly <math>k</math> <math>8</math>s. We then rewrite this more formally, as: | ||
+ | |||
+ | <cmath>x_k=8\biggr(\sum_{n=0}^{k}8(10)^n\biggr)</cmath> | ||
+ | <cmath>=64\biggr(\sum_{n=0}^{k}(10)^n\biggr)</cmath> | ||
+ | <cmath>=64\frac{10^{k+1}-1}{9}</cmath> | ||
+ | |||
+ | Then, finding a recursive formula, we get: | ||
+ | |||
+ | <cmath>x_{k+1}=64\times 10^{k+1}+x_k</cmath> | ||
+ | |||
+ | We will now use induction, Our base case will be <math>k=2</math>. It's easy to see that this becomes <math>x_2=704</math>. Then, the <math>k+1</math> case: let x_k=7111\dots104 with <math>k-2</math> <math>1</math>s. Then <math>x_{k+1}=64000\dots000+7111\dots104</math>. Adding these numbers, we get <math>x_{k+1}=71111\dots104</math>. | ||
+ | |||
+ | Summing these digits, we have <math>4+7+(k-2)=1000</math>, giving us <math>k=991</math>. | ||
==Solution 2(Educated Guesses if you have no time)== | ==Solution 2(Educated Guesses if you have no time)== |
Revision as of 11:31, 2 February 2021
- 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 of each product and is always , so we just do .
Proof of this solution's validity
Since this solution won't fly on a proof-based competition, here's a proof that it's valid:
We will call with exactly s. We then rewrite this more formally, as:
Then, finding a recursive formula, we get:
We will now use induction, Our base case will be . It's easy to see that this becomes . Then, the case: let x_k=7111\dots104 with s. Then . Adding these numbers, we get .
Summing these digits, we have , giving us .
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.