Math Help With Consecutive Integers?

Question

I Need 2 Consecutive Integers Whose Sum Is 116, 2 Even Consecutive Integers Whose Sum Is 126, And 3 Odd Consecutive Integers whose sum is 117. PLZ HELP!!!!!!!!!!!

Answer 1

PROBLEM 1:

The first one is impossible since consecutive integers always sum to make an odd number.

n + n + 1 = 116
2n = 115
n = 57.5

The answer would be 57½ + 58½ = 116 but those aren't integers. Are you sure you stated it correctly?

PROBLEM 2:
First integer = n
Second consecutive integer = n + 2
n + n + 2 = 126
2n + 2 = 126
2n = 124
n = 62

So the answer is 62 and 64 (126)

PROBLEM 3:

Integers:
n
n + 2
n + 4

3n + 6 = 117
3n = 111
n = 37

Odd numbers 37, 39, 41 = 117

Answer 2

If n is an integer, then the integer right after it is going to be (n+1). So "n" and "n+1" are consecutive integers.

If two consecutive integers add up to 116, then you can write this as n + (n+1) = 116. Algebraically solve this equation for n, and you have the smaller of the two integers. The bigger of the two integers is just that integer plus one. Actually, in this case, you'll get 2n+1 = 116, which means 2n=115, or n = 115/2. But this isn't an integer, so there actually isn't a solution.

If an integer is even, then it can be rewritten as 2m for some other integer m. The next even number that comes after 2m would just be that number plus two more (skipping over the odd number), or 2m + 2. So set 2m + (2m+2) = 126, and solve for m. Then use this value to write what 2m and 2m+2 would be.

If an integer is odd, then it's just some even number plus one, so it can be rewritten as 2p+1 for some other integer p. The next odd integer would be this plus two, or 2p+3. The odd integer after that would be another two more, or 2p+5. Add these up, set them equal to 117, and solve for p. Then use p to find 2p+1, 2p+3, and 2p+5.

Answer 3

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