Difference between revisions of "2014 AMC 10A Problems/Problem 20"

Revision as of 00:52, 7 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.

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.

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.

@@ Line 24: / Line 24: @@
 ==Solution 2(Educated Guess if you have no time)==
-We first note that <math>125 \cdot 8 = 1000</math> and so we assume there are <math>125</math> <math>8</math>s. Then we note that it is asking for the second factor, so we subtract <math>1</math>(the original <math>8</math> in the first factor). Now we have <math>125-1=124.</math> The second factor is obviously a multiple of <math>124</math>. Listing the first few, we have <math>124, 248, 372, 496, 620, 744, 868, 992, 1116, 1240, ...</math> 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 <math>\fbox{(D)}</math>. ~mathboy282
+We first note that <math>125 \cdot 8 = 1000</math> and so we assume there are <math>125</math> 8s.
+Then we note that it is asking for the second factor, so we subtract <math>1</math>(the original <math>8</math> in the first factor).
+Now we have <math>125-1=124.</math> The second factor is obviously a multiple of <math>124</math>.
+Listing the first few, we have <math>124, 248, 372, 496, 620, 744, 868, 992, 1116, 1240, ...</math>
+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 <math>\fbox{(D)}</math>. ~mathboy282
 ===Note(Must Read)===

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