Who had what numbers?

Question

Lee chooses a positive integer less than 11 he calls a and an integer called b such that 11 ≤ b ≤ 20. He adds all the positive integers between a and b, including a and b. His sum is 90. Jim does the same as Lee but uses different numbers. He calls them x and y. His sum is also 90. The sum of a and b is greater than x and y. What is a, b, x and y equal to?

Answer 1

The numbers are $$2,3,...,13$$ which add up to $(2+13)(12/2)=90$

We use sum formula for the arithmetic sequence to come up with the answer.

$ n(a+b)/2 = 90$ with all three $n, a,b$ positive integers and $b\ge 11$

Answer 2

$\frac {(a+b)(b-a+1)}{2} = \frac {(x+y)(y-x+1)}{2} = 90$

The source of this formula is probably the trickiest, and most importat part. Can you derive this?

$(a+b)(b-a+1)$ are factors of $180$

$(6,30)$ is one such factor pair.

But if

$b+a = 30\\ b-a+1 = 6\\ 2b +1= 36\\ b = \frac {35}{2}\\ a = \frac {36-6+1}{2}$

If $a, b$ are integers then one of the members of the factor pair must be odd.

$(9,20), (12,15), (5,36), (1,180)$ are the factor pairs of $180$ with one odd factor.

Plugging into the system of equations above.

$(a,b) = (6,14),(2,13), (16,20)$

$(16,20)$ does not fit the required criteria of $a<11, 11\le b \le 20$

nor does $(1,180)$

$(a,b) = (6,14), (x,y) = (2,13)$ fits the final criteria that $a+b > x+y$