2012 AMC 8 Problems/Problem 15

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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 $\boxed{\textbf{(D)}61{ and }65}$

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 (ProblemsAnswer KeyResources)
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

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