I JUST NEED HELP WITH PART B. Please help, THANK YOU!!!! Someone else posted the answer to this question but they said 
but I don't understand how they got that!!
+23251 Since the sum of a + (a + 1) + (a + 2) + ,,, + (a + n - 1) = n(a + (a + n - 1) ) / 2 = n(2a + n - 1) / 2,
we need n(2a + n - 1) / 2 = 100
or n(2a + n - 1) = 200 with both n and a to be whole numbers.
Since n must divide 200, the possible values for n are: 4, 5, 8, 10, 20, 25, 40, 50, 100, 200.
If n = 4: 4(2a + 4 - 1) = 200 ---> 4(2a + 3) = 200 ---> 2a + 3 = 50 ---> impossible
If n = 5: 5(2a + 5 - 1) = 200 ---> 5(2a + 4) = 200 ---> 2a + 4 = 40 ---> a = 18
If n = 8: 8(2a + 8 - 1) = 200 ---> 8(2a + 7) = 200 ---> 2a + 7 = 25 ---> a = 9
If n = 10: 10(2a + 10 - 1) = 200 ---> 10(2a + 9) = 200 ---> 2a + 9 = 20 ---> impossible
and, you can check the rest ...
I realize that this isn't an elegant solution, but ...
+129839 "Brute force" method......LOL!!!
No 2 ,3 or 4 consecutive integer combos are possible
Five consecutive integers 18 - 22 a = 18, n = 5
Six consecutive integers = not possible
Seven consecutive integers = not possible
Eight consecutive integers = 9 - 16 a = 9, n = 8
I think that's all..........
+23251 Since the sum of a + (a + 1) + (a + 2) + ,,, + (a + n - 1) = n(a + (a + n - 1) ) / 2 = n(2a + n - 1) / 2,
we need n(2a + n - 1) / 2 = 100
or n(2a + n - 1) = 200 with both n and a to be whole numbers.
Since n must divide 200, the possible values for n are: 4, 5, 8, 10, 20, 25, 40, 50, 100, 200.
If n = 4: 4(2a + 4 - 1) = 200 ---> 4(2a + 3) = 200 ---> 2a + 3 = 50 ---> impossible
If n = 5: 5(2a + 5 - 1) = 200 ---> 5(2a + 4) = 200 ---> 2a + 4 = 40 ---> a = 18
If n = 8: 8(2a + 8 - 1) = 200 ---> 8(2a + 7) = 200 ---> 2a + 7 = 25 ---> a = 9
If n = 10: 10(2a + 10 - 1) = 200 ---> 10(2a + 9) = 200 ---> 2a + 9 = 20 ---> impossible
and, you can check the rest ...
I realize that this isn't an elegant solution, but ...
