What are consecutive odd numbers have a sum of 75?

Answer 1

There are quite a few possible sets of consecutive odd numbers that sum to 75:

1. with 1 member: {75}
2. with 3 members: {23, 25, 27}
3. with 5 members: {11. 13, 15, 17, 19}
4. with 15 members: {-9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
5. with 25 members: {-21, -19, -17, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27}
6. with 75 members: {-73, -71, -69, ..., 71, 73, 75}

How this was solved.

Let the first odd number be 2n + 1 where n is an integer ≥ 0.

Then a set of m odd integers has members 2n + 1, 2n + 3, ...., 2n + (2m - 1).

This has a sum of (2n + 2n + ... + 2n) + (1 + 3 + ... + (2m - 1)) where there are m terms in each half

= 2mn + m²

So we need a set of m consecutive odd integers with a sum of 75.

m cannot be even as the sum of an even number of odd integers is even; so plug in m = 1, 3, 5, ... and see what value this gives for n, and hence the set:

• m = 1

2mn + m² = 2n + 1 = 75

→ n = 37

→ the set contains {75}

• m = 3

2mn + m² = 6n + 9 = 75

→ n = 11

→ the set contains {23, 25, 27}

• m = 5

2mn + m² = 10n + 25 = 75

→ n = 5

→ the set contains {11, 13, 15, 17, 19}

• m = 7

2mn + m² = 14n + 49 = 75

→ n = 13/7 which is not an integer

→ there is no set of odd consecutive integers with a sum of 75 that has 7 members.

• m = 9

2mn + m² = 18n + 81 = 75

→ n = -1/3 which is not an integer nor is it greater than, or equal to 0

→ there is no set of odd consecutive integers with a sum of 75 that has 9 members.

And so on...

If you notice, the last three sets are the same as the first three sets but with the inclusion of all positive and odd numbers with an absolute value less than the first member of the corresponding set.