2014 AMC 10A Problems/Problem 20

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 1

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 $\boxed{\textbf{(D) } 991}$

Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\boxed{\textbf{(D) } 991}$ .

Proof of Solution 1

Since this solution won't fly on a proof-based competition, here's a proof that it's valid:

We will call $x_k=8(888\dots8)$ with exactly $k$ $8$ s. We then rewrite this more formally, as:

$x_k=8\biggr(\sum_{n=0}^{k-1}8(10)^n\biggr)$ $=64\biggr(\sum_{n=0}^{k-1}(10)^n\biggr)$ $=64\frac{10^{k}-1}{9}$

Then, finding a recursive formula, we get:

$x_{k+1}=64\times 10^{k}+x_k$

We will now use induction, Our base case will be $k=2$ . It's easy to see that this becomes $x_2=704$ . Then, the $k+1$ case: let $x_k=7111\dots104$ with $k-2$ $1$ s. Then $x_{k+1}=64000\dots000+7111\dots104$ . Adding these numbers, we get $x_{k+1}=71111\dots104$ .

Summing these digits, we have $4+7+(k-2)=1000$ , giving us $k=991$ .

Solution 2 (Educated Guessing 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 probably will not work for other problems; it is just a lucky coincidence.

Video Solution

https://youtu.be/wQzuQZvq8sk

~savannahsolver

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.

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