Find the difference of two pronic numbers whose sum is 240. How do I do this without Excel/brute force, and how do I explain this to an 11 year old please? There's apparently supposed to be a trick to this because it's in a standardised exam.
Well I noticed that the difference between two consecutive numbers in the sequence of pronic numbers is an element of $\{4,6,8,10,12,...\}$.
Answer: I guess the pronic numbers are 30 and 210 and hence the difference is 180.
See here for a long solution. I lost my notes, but I think I can use quadratic formula here (or I'm confusing this with another question), but the 11 year old doesn't know quadratic. I think curriculum varies a lot from where she is as opposed to from where I am.

if $a(a+1)+b(b+1)=240$, then must one of them less than $120$ or equal $120$. we can let $a < b$. as $11 \cdot 12 > 120 > 10 \cdot 11$, let us try $A$ less than $10 \cdot 11$, B great than $11 \cdot 12$, here $A=a(a+1), B=b(b+1)$.
the minimum $A$ is $A = 1 \cdot 2$, if $A=1 \cdot 2$, $B$ is the maximum, but if $B = 15 \cdot 16$, we have $A+B=242>240$, so $B < 15 \cdot 16$, the maximum $B$ is $14 \cdot 15$.
so we have $11 \cdot 12 \le B \le 14 \cdot 15$
when $A = 10 \cdot 11$, then if $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$, $A+B \neq 240$. note, when check $A+B \neq 240$, we can try the last digit of $A+B$ $=$ or $\neq$ zero, the same check below.
when $A=9 \cdot 10$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, only need check $B=14 \cdot 15$, but $A+B = 9 \cdot 10+14 \cdot 15 \neq 240$, no solution.
when $A=8 \cdot 9$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A=7 \cdot 8$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A = 6 \cdot 7$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A = 5 \cdot 6$, then we have your solution.
