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)
Let the first 3 consecutive integers be p-1, p, and p+1
So, the next 3 integers are p+2,p+3, and p+4
Then, by the problem:
(p-1+p+p+1)(p+2+p+3+p+4)=111,111,111
or, (3p)*(3(p+3))=111,111,111
or, p(p+3)=111,111,111/9=12345679
Multiplying by 4 and adding 9 to both sides we obtain:
(2p+3)^2= 49382725
or, 2p+3= v(49382725)
or, 2p+3 = 7027.284 (approx)
Then the lhs of the above equation is an integer, while the rhs of the equation is NOT an integer.
This is a contradiction.
Consequently it follows that Bobby DID ERR giving credence to the adage, "To err is human."
A Further Overplus Solution
