Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
 
(26 intermediate revisions by 18 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #16]] and [[2014 AMC 10A Problems|2014 AMC 10A #20]]}}
 
==Problem==
 
==Problem==
  
−
The product <math>(8)(888\dots8)</math>, where the second factor has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>.  What is <math>k</math>?
+
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>The product <math>(8)(888\dots8)</math>, where the second factor has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>.  What is <math>k</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
−
<math> \textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}}\ 991\qquad\textbf{(E)}\ 999 </math>
+
<math>\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999</math>
  
−
==Solution==
+
==Solution 1==
−
Note that <math>8 \cdot \underbrace{888...8}_{k \text{ digits}}=7\underbrace{111...1}_{k-2 \text{ ones}}04</math>, which has a digit sum of <math>7+k-2+0+4=9+k</math>. Since we are given that said number has a digit sum of <math>1000</math>, we have <math>9+k=1000 \Rightarrow k=\boxed{\textbf{(D) }991}</math>
+
We can list the first few numbers in the form <math>8 \cdot (8....8)</math>
 +
 
 +
(Hard problem to do without the multiplication, but you can see the pattern early on)
 +
 
 +
<math>8 \cdot 8 = 64</math>
 +
 
 +
<math>8 \cdot 88 = 704</math>
 +
 
 +
<math>8 \cdot 888 = 7104</math>
 +
 
 +
<math>8 \cdot 8888 = 71104</math>
 +
 
 +
<math>8 \cdot 88888 = 711104</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>\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 Solution 1===
 +
 
 +
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-1}8(10)^n\biggr)</cmath>
 +
<cmath>=64\biggr(\sum_{n=0}^{k-1}(10)^n\biggr)</cmath>
 +
<cmath>=64\frac{10^{k}-1}{9}</cmath>
 +
 
 +
Then, finding a recursive formula, we get:
 +
 
 +
<cmath>x_{k+1}=64\times 10^{k}+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 <math>x_k=7111\dots104</math> 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 Guessing if you have no time)==
 +
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)===
 +
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
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2014|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2014|ab=A|num-b=19|num-a=21}}
 +
{{AMC12 box|year=2014|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Number Theory Problems]]

Latest revision as of 15:37, 4 July 2024

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_20&oldid=221652"