2014 AMC 10A Problems/Problem 20

Revision as of 23:52, 6 December 2020 by Mathboy282 (talk | contribs) (Solution 2(Educated Guess if you have no time))
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 $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$?

$\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999$

Solution

We can list the first few numbers in the form $8 \cdot (8....8)$

(Hard problem to do without the multiplication, but you can see the pattern early on)

$8 \cdot 8 = 64$

$8 \cdot 88 = 704$

$8 \cdot 888 = 7104$

$8 \cdot 8888 = 71104$

$8 \cdot 88888 = 711104$

By now it's clear that the numbers will be in the form $7$, $k-2$ $1$s, and $04$. We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$. Solving, we get $k = 991$, meaning the answer is $\fbox{(D)}$

Solution 2(Educated Guess if you have no time)

We first note that $125 \cdot 8 = 1000$ and so we assume there are $125$ 8s. Then we note that it is asking for the second factor, so we subtract $1$(the original $8$ in the first factor). Now we have $125-1=124.$ The second factor is obviously a multiple of $124$. Listing the first few, we have $124, 248, 372, 496, 620, 744, 868, 992, 1116, 1240, ...$ 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 $\fbox{(D)}$. ~mathboy282

Note(Must Read)

We were just lucky; this method is NOT reliable. Please note that this may not work for other problems.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png