How many positive integers less than 100 can be written as the sum of consecutive positive integers?

Question

a)11
b)9
c)7
d)5

Answer 1

I don't get the question or the answers.

1+2 = 3
2+3 = 5
3+4 = 7
etc.
So I can represent almost half the numbers (3, 5, ..., 99) as the sum of *two* consecutive positive integers.

Continuing with sums of *three* consecutive positive integers, you can eliminate a few more...
1+2+3 = 6
3+4+5 = 12
5+6+7 = 18
etc.

So the number is quite a few and more than all your answers. Could you have left something out of the question such as a certain *number* of consecutive positive integers?

Alternatively, could it mean a sum of consecutive positive integers starting at 1?
e.g.
1+2 = 3
1+2+3 = 6
1+2+3+4 = 10
...
1+2+3+...+12 = 78
1+2+3+...+12+13 = 91

But that is 12 numbers, which again is none of your answers...

Answer 2

None of the above.
All the odd numbers except 1 can. For N odd and not=1, N=(N-1)/2 + (N-1)/2 + 1. That's 49 of them.
Now for the even numbers:
For N even and divisible by 3, N=N/3 - 1 + N/3 + N/3 + 1. Another 16.
For N even and divisible by 5 (but not by 3), N=N/5 - 2 + N/5 - 1 + N/5 + N/5 + 1 + N/5 + 2. Another 6.
For N even and divisible by 7 (but not by 3 or 5), N=N/7 - 3 + N/7 - 2 + N/7 - 1 + N/7 + N/7 + 1 + N/7 + 2 + N/7 + 3. Another 4 (14, 28, 56, 98). You should have the summation formula by now, so I won't repeat it below.
For N even and divisible by 11 (but not by lower primes), we get another 4 (22, 44, 77, 88).
For N even and divisible by 13 (but not by lower primes), we get another 3 (26, 52, 78).
For N even and divisible by 17 (but not by lower primes), we get another 2 (34, 68).
For N even and divisible by 23 (but not by lower primes), we get another 2 (46, 92).
For N even and divisible by 29, 31, 37, 41, 43 and 47 (but not by lower primes), we get another 1 each for a total of 6 more.
Grand total = 49 + 16 + 6 + 4 + 4 + 3 + 2 + 2 + 6 = 82.