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Quantitative Reasoning – Practice – “Tri-factorable” integer!

PSA_PRACTICE_QNR4. 

PSA_PRACTICE_QNR4

Ans: Let the three consecutive integers be (n-1), n & (n+1).

Product= (n-1) * n * (n+1) = n(n²-1).

n(n²-1) < 1000

=> n³-n < 1000

If n= 11, n³-n = 1331-11= 1320. Not possible.

If n= 10, n³-n = 1000-10= 990. Possible & also the highest value of n. In this case the set of the 3 numbers would be 9,10,11 (n-1, n, n+1 respectively).

Number of such sets would be

(1,2,3)

(2,3,4)

(4,5,6)

(9,10,11).

Thus, there will be 9 “tri-factorable” positive integers below 1000. Option (B)

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