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30^72^87 divided by 11 gives reminder ?
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- Fermat little theorem says, [a^(p−1)]/p remainder is 1.
30^10 ,, when divided by 11 remainder is 1.
The unit digit of
72^87 --> 2^87 --> (2^4)^21 * 2^3
now unit place of (2^4)^21 * 2^3 --> 6 * 8 = 48 ---> unit place -> 8
So, 72^87 = 10K + 8
=> 30^72^87 = 30^(10K+8)
30^(10K+8)/11=[(30^10K) * (30^8)]/11= [(30^10K)/11] * (30^8)/11
=1k * (30^8)/11
now, (30^8)/11= [(2 * 11 +8)^8]/11
(8^8)/11=(2^24)/11=[(2^10)^2 * (2^4)]/11=16/11=5 remainder - 9 years agoHelpfull: Yes(7) No(5)
- http://www.quora.com/What-is-the-remainder-of-30-72-87-when-divided-by-11
- 9 years agoHelpfull: Yes(4) No(2)
- 30^72^87=30^6264/11
=(33-3)^6264/11
using binomial theorem,
the last term in the expansion is (-3)^6264/11
=((-3)^2)^3132/11
=9^3132/11
=(11-2)^3132/11
the last term in the expansion is
=(2^(5*626)+2)/11=(32^626)*4/11
((33-1)^626)*4/11=4/11 remainder is 4
- 9 years agoHelpfull: Yes(2) No(8)
- 30^72^87=( 10*3)6264
=10^6264/11 * 3^6264/11
= 1*(3^10)^626*3^4/11
=3^4/11=rem 4 - 9 years agoHelpfull: Yes(1) No(1)
- 5 is the correct answer
- 9 years agoHelpfull: Yes(0) No(1)
- 30/11=8=2^3
(2^3)^72^87/11 =((2)72*3)^87/11=(2^216)^87/11=((2^5)43.2)^87)/11=((32)^(43*87).2^87)/11
=((-1)^odd.(2^5)^17.2^2)/11=((-1)(-1).4)/11=4/11
from remainder theorem
hence remainder=4.. - 9 years agoHelpfull: Yes(0) No(0)
ans. is 5
- 9 years agoHelpfull: Yes(0) No(0)
- Ans 5
72^87=2^87=8 (87/4 gives remainder 3 and 3rd cycle of 2 is 8)
30^8=(11*2+8)^8
=(11*2)^8+8^8
=(11-3)^8
=3^8
=6561/11=5 (no need to take any variable) - 9 years agoHelpfull: Yes(0) No(0)
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