Number Crux

1.N=7!3 How many factors of N are multiple of 10?
(a)736,(b)1008,(c)1352,(d)894.

2.What is the max value of HCF of [n2+1] and [(n+1)2+17]
(a)69,(b)85,(c)170,(d)None.

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Q1 answer

7!
= 2 x 3 x 4 x 5 x 6 x 7
= 24 x 32 x 5 x 7

And 7! 3
= 212 x 36 x 53 x 73

For 10 we need at least one 5 and one 2
Taking one of those factors we get
 211 x 36 x 52 x 73

Total number of factors of the above number
= 12 x 7 x 3 x 4
= 144 x 7 = 1008

 

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