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 "2018 AMC 8 Problems/Problem 21"
 
(35 intermediate revisions by 16 users not shown)
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
−
Looking at the values, we notice that <math>11-7=4</math>, <math>9-5=4</math> and <math>6-2=4</math>. This means we are looking for a value that is four less than a multiple of <math>11</math>, <math>9</math>, and <math>6</math>. The least common multiple of these numbers is <math>11\cdot3^{2}\cdot2=198</math>, so the numbers that fulfill this can be written as <math>198k-4</math>, where <math>k</math> is a positive integer.  This value is only a three digit≤ integer when <math>k</math> is <math>1, 2, 3, 4</math> or <math>5</math>, which gives <math>194, 392, 590, 788,</math> and <math>986</math> respectively. Thus we have <math>5</math> values, so our answer is <math>\boxed{\textbf{(E) }5}</math>
+
Looking at the values, we notice that <math>11-7=4</math>, <math>9-5=4</math> and <math>6-2=4</math>. This means we are looking for a value that is four less than a multiple of <math>11</math>, <math>9</math>, and <math>6</math>. The least common multiple of these numbers is <math>11\cdot3^{2}\cdot2=198</math>, so the numbers that fulfill this can be written as <math>198k-4</math>, where <math>k</math> is a positive integer.  This value is only a three-digit integer when <math>k</math> is <math>1, 2, 3, 4</math> or <math>5</math>, which gives <math>194, 392, 590, 788,</math> and <math>986</math> respectively. Thus, we have <math>5</math> values, so our answer is <math>\boxed{\textbf{(E) }5}</math>.
  
 
==Solution 2==
 
==Solution 2==
−
Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so (z+1) must be mutiples of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of <math>6</math> and <math>9</math>, which means multiples of <math>18(LCM of\ 6, 9)</math>. So let's say <math>z+1 = 18p</math>, then <math>9 \leq z = 18p-1 \leq 90, so </math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5<math>. So our answer is </math>\boxed{\textbf{(E) }5}$ ~LarryFlora
+
Let us create the equations: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of the value of z. Because of <math>6x+2=11z+7</math>, then <math>6x=11z+5=6z+5(z+1)</math>, so <math>(z+1)</math> must be a mutiple of 6. Because of <math>9y+5=11z+7</math>, then <math>9y=11z+2=9z+2(z+1)</math>, so <math>(z+1)</math> must also be a mutiple of <math>9</math>. Hence, the value of <math>(z+1)</math> must be a common multiple of <math>6</math> and <math>9</math>, which means multiples of <math>18 (LCM \text{ of }\ 6, 9)</math>. So, let's say <math>z+1 = 18p</math>; then, <math>9 \leq z = 18p-1 \leq 90</math>, so <math>1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5</math>. Thus, the answer is <math>\boxed{\textbf{(E) }5}</math>.
  
−
==Video Solution==
+
~LarryFlora
−
https://youtu.be/CPQpkpnEuIc - Happytwin
 
  
−
https://youtu.be/PTwMDbsz2xI
+
==Solution 3==
 +
By the [[Chinese Remainder Theorem]], we have that all solutions are in the form <math>x=198k+194</math> where <math>k\in \mathbb{Z}.</math> Counting the number of values, we get <math>\boxed{\textbf{(E) }5}.</math>
  
−
https://youtu.be/7an5wU9Q5hk?t=939
+
~mathboy282
 +
 
 +
==Solution 4==
 +
We can use modular arithmetic. Set up the equations: <math>x \equiv 2 \mod 6,</math> <math>x \equiv 5 \mod 9,</math> and <math>x \equiv 7 \mod 11.</math> These equations can also be written as <math>x+4 \equiv 0 \mod 6,</math> <math>x+4 \equiv 0 \mod 9,</math> and <math>x+4 \equiv 0 \mod 11.</math> Since <math>x+4</math> is congruent to numbers <math>6, 9,</math> and <math>11,</math> then it must also be congruent to their LCM. Thus, <math>x+4 \equiv 0 \mod 198,</math> since 198 is the LCM of <math>6, 9,</math> and <math>11.</math> Since these numbers have to be three digits, they can only be <math>194, 392, 590, 788,</math> and <math>986.</math> This gives us the answer of <math>\boxed{\textbf{(E) }5}.</math>
 +
 
 +
~ethancui0529
 +
 
 +
==Solution 5==
 +
Let <math>N</math> be the three digit positive integer. <math>N = 6a + 2 = 9b + 5 = 11c + 7</math>. Then, we add four to all sides and write <math>N + 4 = 6(a+1) = 9(b+1) = 11(c+1)</math>. Now, we know that <math>N + 4</math> is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so <math>N = 198k - 4</math>. From this, we can figure out that <math>N</math> can be 5 different three digit numbers -- <math>194, 392, 590, 788,</math> and <math>986</math>. <math>\therefore</math>, the answer is <math>\boxed{\textbf{(E) }5}.</math>
 +
 
 +
~DY
 +
 
 +
== Video Solution by Pi Academy (Fast and Easy) ==
 +
 
 +
https://youtu.be/7xlBdxcsP3I?si=lfZZBB7i_6tOlKr-
 +
 
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/kb_0r9fEyEI
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/7an5wU9Q5hk?t=939  
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solutions==
 +
https://youtu.be/CPQpkpnEuIc
 +
 
 +
~ Happytwin
 +
 
 +
https://youtu.be/hoCdk8AC-0c
 +
 
 +
~savannahsolver
 +
 
 +
https://www.youtube.com/watch?v=PjYwbGm_2aM  ~David
  
 
==See Also==
 
==See Also==

Latest revision as of 18:18, 23 November 2024

Problem

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution 1

Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$.

Solution 2

Let us create the equations: $6x+2 = 9y+5 = 11z+7$, and we know $100 \leq 11z+7 <1000$, it gives us $9 \leq z \leq 90$, which is the range of the value of z. Because of $6x+2=11z+7$, then $6x=11z+5=6z+5(z+1)$, so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so $(z+1)$ must also be a mutiple of $9$. Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$, which means multiples of $18 (LCM \text{ of }\ 6, 9)$. So, let's say $z+1 = 18p$; then, $9 \leq z = 18p-1 \leq 90$, so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$. Thus, the answer is $\boxed{\textbf{(E) }5}$.

~LarryFlora

Solution 3

By the Chinese Remainder Theorem, we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{\textbf{(E) }5}.$

~mathboy282

Solution 4

We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent to their LCM. Thus, $x+4 \equiv 0 \mod 198,$ since 198 is the LCM of $6, 9,$ and $11.$ Since these numbers have to be three digits, they can only be $194, 392, 590, 788,$ and $986.$ This gives us the answer of $\boxed{\textbf{(E) }5}.$

~ethancui0529

Solution 5

Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$. Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$. Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$. From this, we can figure out that $N$ can be 5 different three digit numbers -- $194, 392, 590, 788,$ and $986$. $\therefore$, the answer is $\boxed{\textbf{(E) }5}.$

~DY

Video Solution by Pi Academy (Fast and Easy)

https://youtu.be/7xlBdxcsP3I?si=lfZZBB7i_6tOlKr-


Video Solution (CREATIVE THINKING!!!)

https://youtu.be/kb_0r9fEyEI

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=939

~ pi_is_3.14

Video Solutions

https://youtu.be/CPQpkpnEuIc

~ Happytwin

https://youtu.be/hoCdk8AC-0c

~savannahsolver

https://www.youtube.com/watch?v=PjYwbGm_2aM ~David

See Also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AJHSME/AMC 8 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=2018_AMC_8_Problems/Problem_21&oldid=235373"