official quant thread for cat08

[srikar2097]

1)What is the last digit of

1!^1! + 2!^2! + 3!^3! +................100!^100!

Notice that from 5! the last digit contains 0. So the units digit of this entire expression is got from the first 4 digits.
i.e. 1!^1! + 2!^2! + 3!^3! + 4!^4!
=>1 + (2)^2 + (6)^6 + (24)^24

Taking only the last digit in this expansion, for each term.
=>1 + 4 + 6 + 6 = 17

So the last digit is 7. Hope it's clear.

[srikar2097]

2)Find the number of zeros in 1!*2!*3!....*100!


For this, 10 is formed by 5 and 2. Since the no. of 2's in the above is greater than 5's. So we need to find the max. no. of 5's present in each multiple of 5.

in 5!, no. of 5's present =1
10! no. of 5's present =2
15! " = 3
20! " = 4
25! " =6
30! " = 7
35! " = 8
40! " = 9
45! " = 10
50! " = 12
55! " = 13
60! " = 14
…
15 + 16+ 18 + 19 + 20 + 21 + 22 + 24
100! = 25

Adding all these we get 244. So I feel this is the answer.
Any comments?

[rahul_kash]

Originally Posted by srikar2097

2)Find the number of zeros in 1!*2!*3!....*100!


For this, 10 is formed by 5 and 2. Since the no. of 2's in the above is greater than 5's. So we need to find the max. no. of 5's present in each multiple of 5.

in 5!, no. of 5's present =1
10! no. of 5's present =2
15! " = 3
20! " = 4
25! " =6
30! " = 7
35! " = 8
40! " = 9
45! " = 10
50! " = 12
55! " = 13
60! " = 14
…
15 + 16+ 18 + 19 + 20 + 21 + 22 + 24
100! = 25

Adding all these we get 244. So I feel this is the answer.
Any comments?

hey Srikar
I also thought on the same lines , but got confused , because the terms are factorial so , 6! ,7! ,8! , 9! will also gives one 5 each ......

how to tackle these numbers ???

Correct me If I m wrong.......

[shivam_01]

Originally Posted by srikar2097

2)Find the number of zeros in 1!*2!*3!....*100!


For this, 10 is formed by 5 and 2. Since the no. of 2's in the above is greater than 5's. So we need to find the max. no. of 5's present in each multiple of 5.

in 5!, no. of 5's present =1
10! no. of 5's present =2
15! " = 3
20! " = 4
25! " =6
30! " = 7
35! " = 8
40! " = 9
45! " = 10
50! " = 12
55! " = 13
60! " = 14
…
15 + 16+ 18 + 19 + 20 + 21 + 22 + 24
100! = 25

Adding all these we get 244. So I feel this is the answer.
Any comments?

srikar the number of 5s in 100! is 24 rather 25 and also the number of multiples will be also in numbers which are as such not a multiple of 5 such as 6! will have 1 5 and so on

seems the problem is too complicated than what i have understood
jago Quant ke daddu log aur aage bado

[The Last Romeo]

Originally Posted by shivam_01

srikar the number of 5s in 100! is 24 rather 25 and also the number of multiples will be also in numbers which are as such not a multiple of 5 such as 6! will have 1 5 and so on

seems the problem is too complicated than what i have understood
jago Quant ke daddu log aur aage bado

Hi,
This is my approach to the problem
till 4! we'll not have any 5's
from 5! to 9! --> 4 (5's)
from 10! to 14! --> 8(5's)
from 15! to 19! -->12(5's)
'
'
'
from 95! to 99! -->96 (5's)

adding all of them and taking 4 common = 4 * (1+2+3------+19)
= 4 * (19*20)/2
= 4 * 190
= 760
5's in 100! = 24
total 0's should be = 784

[amitkrsingh]

Originally Posted by srikar2097

2)Find the number of zeros in 1!*2!*3!....*100!


For this, 10 is formed by 5 and 2. Since the no. of 2's in the above is greater than 5's. So we need to find the max. no. of 5's present in each multiple of 5.

in 5!, no. of 5's present =1
10! no. of 5's present =2
15! " = 3
20! " = 4
25! " =6
30! " = 7
35! " = 8
40! " = 9
45! " = 10
50! " = 12
55! " = 13
60! " = 14
…
15 + 16+ 18 + 19 + 20 + 21 + 22 + 24
100! = 25

Adding all these we get 244. So I feel this is the answer.
Any comments?

Srikar, rahul and shivam- I agree with all of you. But if you see 5!, 6!, 7!, 8! and 9! will have one 5 each. Hence total no of 5s from 5! - 9! = 5*1

Similarly we can consider nos in group of 5 and each will have same power of 5 like 10!- 14! and so on.

So, modifying srikar's reply total no of zeros will be-

5*244 - 4*24 = 1220 - 96 = 1124.

[srikar2097]

Yup! So as in 5! contains 1 zero. till 9! we have 4 more zeros. Making is 5 in total.

Similarly we see that 5 more contain till the next multiple of 5 i.e. 15!. So 5*244 (got from my earlier post) gives us 1220.

But amit what is the need to subtract 4(4!)? Could you clarify?

[devils_own]

Hi Puys:
Here are 2 Questions:

Might be easy for u guys, but I cudn't understand how to proceed...so help me out!!

Q1. If x = 1+a+a^2+a^3+a^4+..... to infinity (|a|<1),
y = 1+b+b^2+b^3+b^4+.....to infinity (|b|<1),
Then find 1+ab+a^2b^2+a^3b^3+...to infinity

Options:

(a) x+y/(x+y-1) (b) xy/x+y
(c) xy/x+y-1 (d) None of these


Q2.1st 126 natural nos. are put side by side in the ascending order to form a large no:123456...125126. What will be the remainder when this large no is divided by 5625?

Options:
(a) 5126 (b) 126 (c) 26 (d) None of these.

[amitkrsingh]

Originally Posted by srikar2097

Yup! So as in 5! contains 1 zero. till 9! we have 4 more zeros. Making is 5 in total.

Similarly we see that 5 more contain till the next multiple of 5 i.e. 15!. So 5*244 (got from my earlier post) gives us 1220.

But amit what is the need to subtract 4(4!)? Could you clarify?

95!-99! will have 5^22 each. But 100! will have 5^24 and the next group of 5 nos will be 100!-104! but we have only 100! in question. So, i subtracted and added 4*24 to simplify the calculations. Hope that its clear.

[srikar2097]

Thanks, it's clear. But in your explanation I feel it should be 22^5 and not 5^22. Similar for others...