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The question refers to two factors of 132 that are consecutive integers.
The answer is 11 & 12, as 11 x 12 = 132
This can be solved as follows.
Let n be the smaller of the two numbers then (n + 1) is the other number.
n(n + 1) = 132
n2 + n = 132
n2 + n - 132 = 0............which can be factored as (n + 12 )(n - 11) = 0
As we are only concerned with the positive integer result then this occurs when n - 11 = 0, that is when n = 11, thus (n + 1) = 12.
NOTE : Integers can also be negative and this applies to the other solution when n + 12 = 0, so n = -12 and consequently (n + 1) = -11 giving the result that -12 x -11 = 132.
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