[Varun Khullar]

Originally Posted by 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

nagi check again yaar i think its 5..
25 ka euler 20..
17^-1+13^-1/2 for 25 inverse of 17 is 3 and for 13 its 2..
so 3+2 =5.. answer..

[nagi_nov3]

Originally Posted by Varun Khullar

nagi check again yaar i think its 5..
25 ka euler 20..
17^-1+13^-1/2 for 25 inverse of 17 is 3 and for 13 its 2..
so 3+2 =5.. answer..

Originally Posted by 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

i made a simple mistake for got the 2
= 15*2 = 30 mod 25 = 5

buddy i didnt understand this
17^-1+13^-1/2 for 25 inverse of 17 is 3 and for 13 its 2..
could u please explain the concept of inverse

[chetna]

post edited

[reachmonil]

Okies, heres a intersting problem to clear up basics. It was given to me by IdiotR.

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?
2. The number of 'zeroes'?
3. If 50! is written in base 8 system, how many zeroes would it have?

Enjoy!

[varun nakra1]

Issey khelo....32^32^32^32^......uptil infinity..mod 7=?????????????????....

[nagi_nov3]

Originally Posted by reachmonil

Okies, heres a intersting problem to clear up basics. It was given to me by IdiotR.

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?
2. The number of 'zeroes'?
3. If 50! is written in base 8 system, how many zeroes would it have?

Enjoy!

1) last non-zero is 2
2) 10+2 = 12
3) we have to find number of 8's means 2^3
50! will have 2^47 = 4.2^45 = 15 8's

==> 15' 0's will be there

[varun nakra1]

Originally Posted by reachmonil

Okies, heres a intersting problem to clear up basics. It was given to me by IdiotR.

Thee number - 50! (Read 50 factorial)

1. Whats the last non-zero digit?
2. The number of 'zeroes'?
3. If 50! is written in base 8 system, how many zeroes would it have?

Enjoy!

LAst waley ka 15 answer hai..total number of zeroes toh obviously 12 hain..nd for the last nonzero digit..there are three methods..one is to write all the primes nd thier powers that compose 50! then calculate...second is to write 1 to 10 ..11 to 20..nd so on..nd den find the last digit in each 1 to 10 numebrs nd den multiply..third is use Sumit_khanna;s function..he has made a function for the same

[nagi_nov3]

Originally Posted by varun nakra1

Issey khelo....32^32^32^32^......uptil infinity..mod 7=?????????????????....

in sain main bachpan main kela tagoor

4^32^32......... mod 7

EN for 7 = 6

[4^(2^32^32.......)mod 6] mod 7

EN for 6 = 2

so 4^2 mod 7 = 2

[varun nakra1]

Originally Posted by nagi_nov3

in sain main bachpan main kela tagoor

4^32^32......... mod 7

EN for 7 = 6

[4^(2^32^32.......)mod 6] mod 7

EN for 6 = 2

so 4^2 mod 7 = 2

allaay beta..galti nahin kartey khelney mein...u cnt write it as 5^32^32^32..
32^32^32.....mod 7 is 2^5*(32^32.....) mod 7..samjhey

[nagi_nov3]

Originally Posted by varun nakra1

LAst waley ka 15 answer hai..total number of zeroes toh obviously 47 hain..nd for the last nonzero digit..there are three methods..one is to write all the primes nd thier powers that compose 50! then calculate...second is to write 1 to 10 ..11 to 20..nd so on..nd den find the last digit in each 1 to 10 numebrs nd den multiply..third is use Sumit_khanna;s function..he has made a function for the same

what was that function?

guys can anybody solve this

this is the first problem i am unable to do in remainders :(
please help me with this

7.What is the remainder when 2^2001 is divided by 2001?

2001 = 3. 23.29

2^2001 mod 3= 2

2^2001 mod 23
2^22 mod 23 =1 = 23k+1
2^21.2 mod 23 = 24

2^2001 mod 23 =12

2^2001 mod 29 =14

N=3x+2= 23y+12=29z+14

=>N=69k+35=29z+14

is key baad...........?