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Difference between revisions of "2012 AMC 8 Problems/Problem 15"

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==Solution 2(Chinese Remainder Theorem)==
 
==Solution 2(Chinese Remainder Theorem)==
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We have to find the smallest number <math>x</math> for <math>x \equiv 2\mod3</math>, <math>x \equiv 2\mod4</math>, and <math>x \equiv 2\mod5</math>. We are excluding six because it won't make our divisors relatively prime. For the first equation, <math>x \equiv 2\mod3</math> holds remainder(<math>r</math>) of 2. Now we have to find <math>M</math>. Our <math>M</math> which is all divisors multiplied by each other divided by the divisor we are using. So our <math>M</math> is <math>\frac{3 \cdot 4 \cdot 5}{3}=20</math>. We now need to find the inverse of <math>a</math> times <math>M</math> and solve <math>a</math>. Doing this we can get <math>20a \equiv 1\mod3</math>, and the smallest value of <math>a</math> where this is true is <math>2</math>. Now we multiply every value we have which gets us <math>2 \cdot 20 \cdot 2</math>. Then we repeat for mod 4 and mod 5. For mod 4 the numbers <math>r</math>,<math>M</math>,<math>a</math> are <math>2</math>, <math>15</math>, and <math>3</math> respectively. For mod 5 we get <math>2</math>,<math>12</math>, and <math>8</math>. Now we have to multiply all our values and we get <math>80</math> (mod 3), <math>90</math> (mod 4), and <math>192</math> (mod 5). Now we do <math>80+90+192\mod{5 \cdot 4 \cdot 3}</math>. This is <math>362\mod60</math>. This means <math>x \equiv 2\mod60</math>, but since it needs to be greater than <math>2</math>, we get <math>62</math>, which divided by six also has a remainder of 2. So <math>62</math> is in the range of <math>\boxed{\textbf{(D)}61 and 65}</math>
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We have to find the smallest number <math>x</math> for <math>x \equiv 2\mod3</math>, <math>x \equiv 2\mod4</math>, and <math>x \equiv 2\mod5</math>. We are excluding six because it won't make our divisors relatively prime. For the first equation, <math>x \equiv 2\mod3</math> holds remainder(<math>r</math>) of 2. Now we have to find <math>M</math>. Our <math>M</math> which is all divisors multiplied by each other divided by the divisor we are using. So our <math>M</math> is <math>\frac{3 \cdot 4 \cdot 5}{3}=20</math>. We now need to find the inverse of <math>a</math> times <math>M</math> and solve <math>a</math>. Doing this we can get <math>20a \equiv 1\mod3</math>, and the smallest value of <math>a</math> where this is true is <math>2</math>. Now we multiply every value we have which gets us <math>2 \cdot 20 \cdot 2</math>. Then we repeat for mod 4 and mod 5. For mod 4 the numbers <math>r</math>,<math>M</math>,<math>a</math> are <math>2</math>, <math>15</math>, and <math>3</math> respectively. For mod 5 we get <math>2</math>,<math>12</math>, and <math>8</math>. Now we have to multiply all our values and we get <math>80</math> (mod 3), <math>90</math> (mod 4), and <math>192</math> (mod 5). Now we do <math>80+90+192\mod{5 \cdot 4 \cdot 3}</math>. This is <math>362\mod60</math>. This means <math>x \equiv 2\mod60</math>, but since it needs to be greater than <math>2</math>, we get <math>62</math>, which divided by six also has a remainder of 2. So <math>62</math> is in the range of <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>.
  
 
If CRT didn't make sense, this video is very good:
 
If CRT didn't make sense, this video is very good:

Latest revision as of 14:08, 16 August 2025

Problem

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

$\textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99$

Video Solution

https://youtu.be/rQUwNC0gqdg?t=172

https://www.youtube.com/watch?v=Vfsb4nwvopU ~David

https://youtu.be/hOnw5UtBSqI ~savannahsolver

Solution

To find the answer to this problem, we need to find the least common multiple of $3$, $4$, $5$, $6$ and add $2$ to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. $3 = 3^{1}$, $4 = 2^{2}$, $5 = 5^{1}$, and $6 = 2^{1}*{3^{1}}$. So the least common multiple of the four numbers is $2^{2}*{3^{1}*{5^{1}}} = 60$, and by adding $2$, we find that that such number is $62$. Now we need to find the only given range that contains $62$. The only such range is answer $\textbf{(D)}$, and so our final answer is $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

~NXC

Solution 2(Chinese Remainder Theorem)

We have to find the smallest number $x$ for $x \equiv 2\mod3$, $x \equiv 2\mod4$, and $x \equiv 2\mod5$. We are excluding six because it won't make our divisors relatively prime. For the first equation, $x \equiv 2\mod3$ holds remainder($r$) of 2. Now we have to find $M$. Our $M$ which is all divisors multiplied by each other divided by the divisor we are using. So our $M$ is $\frac{3 \cdot 4 \cdot 5}{3}=20$. We now need to find the inverse of $a$ times $M$ and solve $a$. Doing this we can get $20a \equiv 1\mod3$, and the smallest value of $a$ where this is true is $2$. Now we multiply every value we have which gets us $2 \cdot 20 \cdot 2$. Then we repeat for mod 4 and mod 5. For mod 4 the numbers $r$,$M$,$a$ are $2$, $15$, and $3$ respectively. For mod 5 we get $2$,$12$, and $8$. Now we have to multiply all our values and we get $80$ (mod 3), $90$ (mod 4), and $192$ (mod 5). Now we do $80+90+192\mod{5 \cdot 4 \cdot 3}$. This is $362\mod60$. This means $x \equiv 2\mod60$, but since it needs to be greater than $2$, we get $62$, which divided by six also has a remainder of 2. So $62$ is in the range of $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

If CRT didn't make sense, this video is very good:

Chinese Remainder Theorem | Sun Tzu's Theorem

Additionally, this page is great:

CRT

~AM24

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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