Prove that any sum sequence of 6 integers is divisible by 6
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First get the prime factors of 6. These are 3 and 2, once each. 6 = 3*2.
To be divisible by 6, a number needs to be divisible by both 2 and 3.
So how about 3 consecutive numbers, meaning 3 consecutive integers? You can write these as n-1, n and n+1, where n is the middle number.
Now n is either even or odd. If it is odd, then n-1 and n+1 are even. Even means divisible by 2, so the product of three consecutive integers has to be divisible by 2.
Consider n again. If you divide it by 3, the remainder will be 0, 1 or 2. If 0, n is a multiple of 3. If 1, it is one greater than a multiple of 3, which would be n-1. If 2, it is two greater, which means it is 1 less than the next multiple of 3, which would be n+1. So one of n-1, n and n+1 has to be a multiple of 3.
What we have shown is that between them, n-1, n and n+1 include a multiple of 2 and a multiple of 3. This means that both 2 and 3 divide their product, which in turn means that their product is a multiple of 6.
To be divisible by 6, a number needs to be divisible by both 2 and 3.
So how about 3 consecutive numbers, meaning 3 consecutive integers? You can write these as n-1, n and n+1, where n is the middle number.
Now n is either even or odd. If it is odd, then n-1 and n+1 are even. Even means divisible by 2, so the product of three consecutive integers has to be divisible by 2.
Consider n again. If you divide it by 3, the remainder will be 0, 1 or 2. If 0, n is a multiple of 3. If 1, it is one greater than a multiple of 3, which would be n-1. If 2, it is two greater, which means it is 1 less than the next multiple of 3, which would be n+1. So one of n-1, n and n+1 has to be a multiple of 3.
What we have shown is that between them, n-1, n and n+1 include a multiple of 2 and a multiple of 3. This means that both 2 and 3 divide their product, which in turn means that their product is a multiple of 6.
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