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Let three consecutive integers be n, n+1 and n+2.
If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2.
If n is not divisible by 3 then it will leave a remainder of 1 or 2.
If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3.
If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
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How do you prove that the product of two consecutive even integers is divisible by 8?
Because 6*8 = 48 and 48/8 = 6
Prove that the two consecutive positive integer is divisible by 2?
There is something missing from your question: what about the two consecutive integers? Is the proof required that one of them is divisible by 2? Or that their product is (which amounts to the same thing)? Showing that one of two consecutive number is divisible by 2: Suppose your two numbers are n and (n+1). If n is divisible by 2, ie n = 2k, the result is shown. Otherwise assume n is not divisible by 2. In this case n = 2m+1. Then: (n+1) = ((2m+1)+1) = 2m + 2 = 2(m+1) which is a multiple of 2 and so divisible by 2. QED. Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part: "as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2." To show the product is divisible by 2, show either n is divisible by 2 or (n+1) is as above, then the result follows as one of n and (n+1) is divisible by 2 and so their product is.
How do you prove 2n is the sum of two odd consecutive integers?
2n can be split into 2 n's so: n+n then add one to one of the n's and subtract one from one of the n's n+1+n-1 ^two consecutive odd integers^
The sum of 25 consecutive integers is divisible by?
The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!
How do you find the product of consecutive integers?
Call the two consecutive integers n and n+1. Their product is n(n+1) or n2 +n. For example if the integers are 1 and 2, then n would be 1 and n+1 is 2. Their product is 1x2=2 of course which is 12 +1=2 Try 2 and 3, their product is 6. With the formula we have 4+2=6. The point of the last two examples was it is always good to check your answer with numbers that are simple to use. That does not prove you are correct, but if it does not work you are wrong for sure!
