Factorials
Q.1 If p ≥ 5, K - B = 9, B- C= 24 and C= 5! + 6! +7!+ …………+ p! then which is not true for K.
(1) K is a composite no.
(2) K is an odd no.
(3) K is a perfect square
(4) K is divisible by 3
ANS. Let p=5 then K= 153 . Option (3)
Q.2 A = 132x226 divide completely B! then least value of B is
(1) 26
(2)13
(3)30
(4) 39
ANS. Clearly n! contain two 31 and twenty six - 2, therefore least value of n is 30!.
Q.3 How many 17 are there in 100! x 49! ?
(1) 8
(2)7
(3)22
(4) 24
ANS. Clearly 17 are 5 times in 100! and 2 times in 49!. Therefore total 5 times in100! x 49!
Q.4 How many zeros are there in the end of 100! x 49! ?
(1) 18
(2)25
(3)34
(4) 24
ANS. Check only 5 as 5 x 2 gives one zero and we know that highest prime number occur less
than smallest prime. Clearly 5 are 24 times in 100! And 10 times in 49!. Therefore total 34 times in
100! x 49! . Option (3)
Q.5 S =100! x 49! + 2892x 17 + 85x342x289, if we divide S+17 by 175 then remaider will be?
(1) 17
(2)172
(3) 174
(4) 175
ANS. Clearly S divide by 175. Therefore remainder will be 17.
Q.6 If p ≥ 10, A - B = 9, B- C= 144 and C= 6! +7!+ ……+ p! .Find remainder when A divide by 5.
(1) 4
(2) 2
(3) 3
(4) 0
ANS. Let p=10 then A= 1!+2!+…..10! = (…..13) . Clearly Option (3) as when number divide by 5 unit
digit will be the remainder
Q.7 @ - % = 9, % - # = 144 and # = 6! +7!+ 40320 + 9! + 3628800 + 11! + ….25! then unit digit of @
is
(1)
(2) 1
(3) 2
(4) 9
ANS. 1! =
1
2! =
4
3! =
6
4! =
24
5! =
120
6! = ….. 0 (as we get zero on unit place then no need to solve further.)
.
.
25! = …….0
On adding = ….....3 , Unit digit is 3
Clearly Option (1)
Q.11 If, A =10! + 12! + 14!…………+ 210! And B = 11! + 13! +…..99! then hundred digit of 'A + B
+ 200' must divisible by
(1) 4
(2) 1
(3) 2
(4) all of these
ANS.. 10! = 3628800
11! =
….800
12! = .....600
13! =
….800
14! =
….200
15! =
….000
.
.
(as we get zero on hundred place then no need to
.
.
solve further.)
210! =
….000
On adding .......200
Therefore hundred digit of A + B is 2 and hundred digit of A + B + 200 is 4. Option (1)
Q.12 If 3(n-1) x (n-3)! = n! where n3 > 9 and n is natural number. Then value of (n!)n:
(1) 27
(2) 216
(3)256
(4) 331776
3
ANS. clearly n>2 as n >9, put n=3 in question, we get 6=3(2)(1) ,satisfied.
therefore, (n!)n =63 = 216. Clearly option (2)
Q.13 If (n+2)! = 2550 n!, find n.
(1) 49
(2) 216
(3)256
(4) 331776
ANS. on expanding we have, (n+2)(n+1) = 2550,
Now, LHS= two consecutive term, therefore RHS may be two consecutive term.
=> (n+2)(n+1) =50 x 51
=> on comparing (n+1) = 50 and n = 49
Q.14 If (n+1)! = 60 (n-2)!, find n.
(1) 27
(2) 216
(3) 4
(4) 33
ANS. on expanding we have, (n+1)n(n-1) = 60,
Now, LHS= three consecutive term, therefore RHS may be three consecutive term.
=> (n+1)n(n-1) = 5 x 4 x 3 , => on comparing, n = 4
Q.15 Find remainder when n555555 divide by 31 where n can be obtained by the equation (n+3)! = 60 n!
(1) 0
(2) 1
(3) 11
(4) 4
ANS. n=2, therefore 2555555 or 32111111divide by 31. Clearly remainder is 1
Q.16 A= 1! n + 2! n-1 + 3! n-2….+ n!1 , find unit digit of 'A' if n must be the least multiple of 9 and 11.
(1) 1
(2) 3
(3) 6
(4) 7
ANS. A =...1+....4+…6+ …6+…….0+….0+…. all other have unit digit = 0
A = …….7
( all unit digits can be easily find through unit digit method)
Q.17 For n>0, (2n)!/(n!)2 is divisible by
(1) n+1
(2) n+5
(3) 4n+5
(4)4n-1
ANS.Put n= 1 we get (2n)!/(n!)2 =2, clearly option (1)
Put n= 2 we get (2n)!/(n!)2 = 6, clearly option (1)
Put n= 3 we get (2n)!/(n!)2 = 20, clearly option (1)
Q.18 If, A =1! + 2! + 3! +…………+ 123450! , and if A=B, find remainder when AB
divided by 5 ?
(1) 1
(2)10
(3) 0
(4) 7
(……13)
ANS. A= (…….13)
Clearly unit digit of A= (…………3)(….13)
unit digit of A= (…………3)1= 3 according to the cyclic order of three.
Clearly remainder =3 when devided by 5
Q.19 Find remainder when 3(14!)-15(13!) devides 28(13!)+ 27!
(1) 13!
(2)10!
(3)8!
(4)12!
ANS.
(28(13!)+ 27!) / 3(14!)-15(13!)
= ( 28(13!)+ 27!) / 13! (27)
= 28(13!) / 13! (27)
+ 27! / 13! (27)
= (27+1)13! / 13! (27) + 27! / 13! (27)
= (27) 13! / 13! (27) + 13! / 13! (27) + 27! / 13! (27)
Clearly 1st and 3rd are divisible and therefore from 2nd term remainder= 13!
Q.20 If N! has 100 zeros in the end. Find least value of N ?
(1) 278
(2)338
(3) 490
(4) 405
ANS. Trick multiply 100 by 4 to get nearest number to N. therefore 400! But in 400! we get
only 99 zeros. Therefore smallest possible value of N is 405.
Practice questions
Q.1
If, A= C! + (C+2)! + ….. And B = D! + (D+2)! +….. where C and D are even number greater
than 9 and odd number greater than 10 respectively then
(i) Find hundred digit of 'A + B + 28 '
(1) 3
(2) 1
(3) 2
(4) 4
(ii) Last three digit of 'A + B + 28 ' must not divisible by
(1) 3
(2) 1
(3) 2
(4) 4
(iii) Find last three digit of Z, where Z = A + B + W – 3628800 and W = 1! + 2! + 3! +.…+ 10!
(1) 123
(2) 313
(3) 321
(4) none
(iv) If ZZ devide by 5, remainder will be
(1) 3
(2) 1
(3) 2
(4) 4
(v) T = product of last three digit of Z, where Z = A + B + W – 3628800 and W = 1! + 2! + 3! +.…+
10! then which must not be divisible by T.
(1) 126
(2) 243
(3) 321
(4) 513
Q.2 If (n+1)! = 12 (n-1)!, find n.
(1) 3
(2) 1
(3) 2
(4) 4
Q.3 If (n-10)! = 210 (n-13)!, find n.
(1) 31
(2) 17
(3) 12
(4) 4
Q.4 If (n-10)! = 210 (n-13)!, Find remainder when 2881122334455 divide by n2
(1) 0
(2) 1
(3) 288
(4) 4
Q.5 Find remainder when n500divide by 17 where n can be obtained by the equation (n+1)! = 60 (n-2)!
(1) 0
(2) 1
(3) 288
(4) 4
Q.6 A = 132x226 divide completely B! then value of B can be
(1) 26
(2)13
(3)34
(4) none
Q.7 g! divisible by 412 x341 then sum of digits of possible value of ‘g’ is
(1) 26
(2)13
(3)12
(4) none
Q.8 A = 1! n + 2! n-1 + 3! n-2…………+ n!1 , where n which is a two digit number must be the multiple
of 9 and 11. If ‘p’ is unit digit of 'A', then p2244 give remainder 1when divide by
(1) 50
(2) 48
(3) 25
(4) all can be the answer
Q.9 A = 172x233 divide completely B! then minimum possible value of B can be
(1) 34
(2)17
(3)36
(4) 38
Q.10 Find the number of zeros in the end of the following
(1) 100! × 100!
(2) 100! × 12! + 32!22
(3) 100!10 + 42!240
(4) 100! × ( 1! +2! +…. .100! )
(5) 124! × ( 100! + 2525 23!)
(6) (4!+ 5! ….100!) ( 1! +2! +…. .100! )
(7) 222111× 5252 + 7!5! × 18!6!
(8) ( 1! +2! +…. .100! ) 100!
(9) 650! + 10!
(10) 25100! × 100!
Answers
1. (i) (4)
2. (1)
3. (2)
(ii) (1)
(iii) (2)
(iv) (3)
(v) (3)
4. (3)
5. (2)
6. (3)
7. (3)
8. (4)
9. (3)
10. (1)48
(9)2
(2)26
(10)97
(3)240
(4)24
(5)50
(6) no zero
(7) no zero (8) no zero