Bobby found the sum of 3 consecutive integers
and then the sum of the next 3.
Multiplying both results he got 111111111.
Is it possible?
Source Russian Kвант (Quant)
What Bobby did is calculate some number N by evaluating the expression [x+(x+1)+(x+2)]*[(x+3)+(x+4)+(x+5)] for some integer x.
A bit of simplification gives 9x^2+45+36=N. Multiplying each side by 4 will allow us to complete the square and stay within the domain of integers: (6x+15)^2 = 4N+81.
Bobby's result was N=111111111. The prior equation suggests for Bobby to be right then 4*111111111+81=444444525 is a perfect square. sqrt(444444525)=21081.853, thus Bobby did indeed err.
Another solution (overkill?)
