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Winter 2016 (Combinatorics) AMC10 Homework solutions
I'm Fan, I can't quite understand P10's solution:
P10. How many ways are there to arrange the numbers 11,21,31,41,51,61 so that the sum each consecutive group of 3 numbers is divisible by 3?
Answer 3!·(2!· 2!· 2!) = 48.
Solution
As before, first consider our numbers mod 3. Note that the first triple
must be same as the second triple (why?) and hence the first triple
(and second triple) must consist of 0,1,2. Note any such arrangement
works. Afterwards we arrange the actual numbers.
What do it mean by "mod 3".
Hello Fan,
"$\text{mod} 3$" is a Number Theory concept that refers to the remainders of the numbers when they are divided by $3$. So, since $11 = 3\times 2 + 2$, we say that $11$ is the same as $2$ $\text{mod} 3$. So, the numbers $11$, $21$, $31$, $41$, $51$ and $61$ are the same as $2$, $0$, $1$, $2$, $0$ and $1$ $\text{mod} 3$.
In this problem we only need to look at the remainders when we divide by $3$, so modular arithmetic is quite useful.
D
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