2012 AMC 8 Problems/Problem 15

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

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2012 AMC 8 Problems/Problem 15

Contents

Problem

Video Solution

Solution

Solution 2(Chinese Remainder Theorem)

See Also