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what will be the remainder when 30^72^87 is divided by 11.
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- 30^72^87 mod 11
30^72^87 mod 11
Euler value of 11 = 10
72^87 mod 10
(7*10+2)^87 mod 10
2^1 mod 10 = 2
2^2 mod 10 = 4
2^3 mod 10 = 8
2^4 mod 10 = 6
2^5 mod 10 = 2 [Cyclicity is 4... Rem = (4k+3)rd value]
2^87 mod 10 = 8
Now 30^8 mod 11
(11*2+8)^8 mod 11
8^8 mod 11
2^24 mod 11
Euler value of 11 = 10
(2^10)^2 * 2^4 mod 11
16 mod 11 = 5
Ans : 5
- 10 years agoHelpfull: Yes(19) No(0)
- Fermat little theorem says, (a^(p−1))/p remainder is 1.
ie., 30^10 or 8^10when divided by 11 remainder is 1.
The unit digit of 72^87 is 8 (using cyclicity of unit digits)
So 72^87 = 10K + 8
(30^(10K+8))11=(((30^10)^K).30^8)/11=(1^k).(30^8)/11
(8^8)/11=(2^24)/11=16/11=5(remainder) - 10 years agoHelpfull: Yes(2) No(0)
- N = a^x*b^y*c^z* ... etc
Euler value of N is => N*(1-(1/a)) * (1-1/b) * (1-1/c)..etc
Lets take a number 10
10 => 2*5
E(10) = 10*(1-1/2)*(1-1/5) => 4
Lets take 21
21 => 3*7
E(21) = 21*(1-1/3)*(1-1/7) => 12
Hope you understand this @sudhir sharma
- 10 years agoHelpfull: Yes(2) No(0)
- Lets take an example...
If we want to calculate remainder of 2^17 divided by 10
Can u say the value of 2^16 without calculator? ... Its impossible na...
2^1 mod 10 = 2
2^2 mod 10 = 4
2^3 mod 10 = 8
2^4 mod 10 = 6
2^5 mod 10 = 2
2^6 mod 10 = 4
2^7 mod 10 = 8
2^8 mod 10 = 6
[Note here u dont need to calculate 2^8... 2^8 is 2^7*2^1...
So remainder of 2^7 * 2^1=8*2 = 16 mod 11 = 6]
If u observer carefully, after 4th power, the remainder is again 2,4,8,6... Again it goes on like 2,4,8,6...[This is cyclicity]
So remainder of 2^16 ll be 4k+n since 1 cycle get over upto 4th power...
17 => 4k+n => 4*4 + 1
So remainder of 2^17 mod 10 ll be 2^n = 2^1 = 2...
Hope u understand this... @Suraj
- 10 years agoHelpfull: Yes(2) No(0)
- saraswathy, how u calculate euler's value?
- 10 years agoHelpfull: Yes(0) No(0)
- Thanks @ Saraswathy
- 10 years agoHelpfull: Yes(0) No(0)
- how saraswaty???i could not understand...
- 10 years agoHelpfull: Yes(0) No(0)
- What u cant understand @Suraj jha...
- 10 years agoHelpfull: Yes(0) No(0)
- what about cyclicity???i could not understand...
- 10 years agoHelpfull: Yes(0) No(0)
- if you dont have any problem....can u please tell the answer on the email id i just sent to you...
- 10 years agoHelpfull: Yes(0) No(0)
- a bit...problem remains
- 10 years agoHelpfull: Yes(0) No(0)
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