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If (n + 2)!/n! = 156, what is the value n ?

A. 2/131
B. 9
C. 10
D. 11
E. 12

the left hand side evaluates to (n+2)(n+1) = 156

If you imagine N such that N = n+1, then this expression becomes N*(N+1)=156

Or

Product of two consecutive numbers must be equal to 156. From the option choices -

A - 2/131 and 1 + 2/131 will never give 6 in the units digit
B - product of 9 and 10 (which is N+1 here) does not give 6 in the units digit
C - product of 10 and 11 (which is N+1 here) does not give 6 in the units digit
D - rejected on the same grounds
E - 12 (Bingo, 12 and 13 does give 6 in the units digit)

So, N must be 12, which means n+1 = 12 or N = 11
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(n+2)!/n!= (n+2)(n+1)(n!)/n!=(n+2)(n+1)=156

Factorise 156.
156= 13*12

(11+2)*(11+1)=156
n=11

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