[nagi_nov3]

Originally Posted by kingnitin

What is the remainder when 4(4!) + 5(5!) + 6(6!) + .....................+ 19(19!)+20(20!)
is divided by 64?

Try this out!...pls give approaches to the problem

4.4!= (5-1)41= 5!-4!
similarly all will gets caneclled
apart from 5 n 21

5! +21! mod 64 = 120 mod 64 = 56

[kingnitin]

Originally Posted by nagi_nov3

4.4!= (5-1)41= 5!-4!
similarly all will gets caneclled
apart from 5 n 21

5! +21! mod 64 = 120 mod 64 = 56

Thnaks for the approach...but. how does 4! get cancelled??

(5*4! - 4!) + (6*5! - 5!) + (7*6! - 6!) +.......................+ 20*19! - 19! + 21*20! - 20!

(5! - 4!) + (6! - 5!) + (7! - 6!) + (8! -7!) +...................................+(20! - 19!) + (21! - 20!)

Tis leaves

( - 4! + 21!) MOD 64

= - 4! (MOD) 64 + 0

= - 24

So Rem = 64 - 24 = 40

[nagi_nov3]

Originally Posted by kingnitin

how does 4! get cancelled??

(5*4! - 4!) + (6*5! - 5!) + (7*6! - 6!) +.......................+ 20*19! - 19! + 21*20! - 20!

(5! - 4!) + (6! - 5!) + (7! - 6!) + (8! -7!) +...................................+(20! - 19!) + (21! - 20!)

Tis leaves
5! - 4! + 21!...???

mistake

it leaves -4! +21! only even 5! gets cancelled

so -24 = 40

[ashish banaya..]

Originally Posted by kingnitin

What is the remainder when 4(4!) + 5(5!) + 6(6!) + .....................+ 19(19!)+20(20!)
is divided by 64?

Try this out!...pls give approaches to the problem

Me getting 40 as the ans.
Here we need to care only till 7*7! b'coz after that it is divisible by 64
Therefor rem. are 32+24+32+16=40

[kingnitin]

Originally Posted by ashish banaya..

Me getting 40 as the ans.
Here we need to care only till 7*7! b'coz after that it is divisible by 64
Therefor rem. are 32+24+32+16=40

U are right ..ashish..silly mistake I have edited my post
ans is 40

[drizzle]

Originally Posted by ashish banaya..

Me getting 40 as the ans.
Here we need to care only till 7*7! b'coz after that it is divisible by 64
Therefor rem. are 32+24+32+16=40

<Edit> i got it where i did mistake...

[warrior]

Originally Posted by kingnitin

What is the remainder when 4(4!) + 5(5!) + 6(6!) + .....................+ 19(19!)+20(20!)
is divided by 64?

Try this out!...pls give approaches to the problem

4(4!) = 5! -4!
ans so on for all terms

you will be left with 21!-4!

21! is divisible by 64

so rem is -24 or 40

[kabir1224]

Boss me also getting the ans to be 40? guys got a ques can v people devise some method to find the sum of series of factorials for ex 3!+4!+5!+.....+n!? got any bright ideas please do put forth

[nagi_nov3]

Remainders Remainders
lets solve problems posted by Varun
i tried some and doing rest

1.What is the remainder when 17^19 + 13^19 is divided by 25?
(15+2)^19 +(15-2)^19
=2.19.15.2^18 mod 25
= 1.19.3.(64)^3 mod 5 (*5)
=-57 mod 5 (*5)
=3*5 = 15

2.Find the remainder when 104^303 is divided by 101.
101 euler no. is 100

so 104^300 = 1
104^3 mod 101 = 3.3.3 = 27

3.Find the last two digits in the expansion of 2^ 999.

2^999 mod 100 = 2^997 mod 25 (*4)
E.N for 25 = 20

=2^17 mod 25(*4)
=1024.128 mod 25(*4)
=-1.3 mod 25(*4)
=22 (* 4) =88

4.Find the remainder when 2 ^ 1990 is divided by 1990.

N=2^1989 mod 995 (*2)

since 995 =199*5

2^1989 mod 5 = 2
EN for 199 = 198
2^1989 mod 199 =2^9 mod 199
512 mod 199= 14

remainder=5K+2= 199L+14
5k = 199L+12
l=2
so R=512

so N= 995a+512 mod 995 = 512

so final remainder is 1024


5.Remainder when (128 )^500 is divided by 153.
En for 153 = 96

128^20 = -25^20 = 625^10 = 13^10 = 16^5 = 103. 103.16 = 10609.16 = 67

or

153 =17.9
128^20 mod 9 = 2^20 mod 9 = 4
128^20 mod 17 = 9^20 mod 17 = 3^40 mod 17 = 3^8 = 10.10.9 = -18 = 16
N= 9x+4 = 17y+16
9x = 17y +12 => y = 12
N= 220
mod 153 = 67


6.Find last 3 digits of 3^1994.

3^1994 mod 1000 =
EN for 1000 = 400

3^394.3^6 = 001
this is one way
3^394.729 = 001

or

9^997 mod 1000
(10-1)^997
= 997c2 . 100. (-1)^995+997c1.10.(-1)^996+(-1)^997

= - 997c2.100+9970 -1
= 3. 800 +970-1
=400+970-1
=369

try from here

7.What is the remainder when 2^2001 is divided by 2001?
8.What is the remainder when 2^2002 is divided by 2002?
9.What is the remainder when 13^2404 is divided by 2310?
10.What is the remainder when 2^10013 is divided by 3125?
11. Find last 4 digits of (2319)^{10^12 + 2}.
And a lengthy one here.
12. What is the remainder when 17^28820 is divided by 30030?

[warrior]

Originally Posted by nagi_nov3

9.What is the remainder when 13^2404 is divided by 2310?

2310 ka eulker hai 480

so we are left with 13^4

so now multiply and divide and we get 841