Solution (b)1The unit digit of every term from 5! to 25! is 0.

Also, 1! – 2! + 3! – 4! = 1 – 2 + 6 – 24 = –19.

Hence, the unit digit of N will be 10 – 9 = 1. The unit digit of N^{2524} will also be 1.

Solution (a) 0

We have to calculate the number of zeroes starting from the right end of the number N. The number of zeroes from:

1! to 4! = 0

5! to 9! = 1× 5 = 5

10! to 14! = 2 × 5 = 10

15! to 19! = 3 × 5 = 15

20! to 24! = 4 × 5 = 20

25! to 29! = 6 × 5 = 30

30! to 34! = 7 × 5 = 35

35! to 39! = 8 × 5 = 40

So we get 155 zeroes till 39! only. From this we can easily conclude that the 152^{nd} digit from the right end of N will be zero.

Solution (a) 37! + 36! – 2

Let E = 0!(1²+1-1) + 1!(2²+2-1) + 2!(3²+3-1) + 3!(4²+4-1) + …………+ 35!(36²+36-1)

On expanding the first two terms of the series, we get

1×1! + 1! – 1×0! + 2×2!+ 2! -1! + 3×3! + 3! -2! +…….36×36! +36! – 35!

= 1! (1+1) -1×0! + 2!(2+1) – 1! + 3!(3+1) -2!+………..+ 36!(36+1) -35!

= 2! -0!+3!-1!+4!-2!+…………+37!-35!

= 2! -0!+3!-1!+4!-2!+…………+37!-35!

Note that the value of 0! Is 1

= 37!+36!-2

Sol. a)
Option a – 33! = 12!×(13×14×15×16×………×33)Option b – 10!×13! = 12! ×(13×10!)

Option c – 12!×11! = 12!× (11×10!)

Option d – 30!×3! = 12!×(3!× 13×14×15………..×30)

Now, clearly a>d>b>c

Thus option a has the maximum value.

Penultimate question is having some printing mistake please correct it, a 2 is extra printed there. Thanks&Regards.

It was a typographical error that has been corrected. Thank you for pointing it out. 🙂