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17 Sep 2024, 05:30
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We need to find the remainder when 333^222 is divided by 7?
We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.
=> \(333^{222}\) = \((329 + 4)^{222}\)
Watch this video to MASTER BINOMIAL Theorem
Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 222C222 * 329^0 * 4^222) is divided by 7
=> Remainder of 4^222 is divided by 7
=> Remainder of 2^444 divided by 7
To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7
Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1
=> Cycle is 3
Remainder of 444 by 3 = Remainder of (4+4+4) by 3 = 0
=> Remainder of \(2^{444}\) by 7 = Remainder of \(2^{Cycle}\) by 7 = Remainder of \(2^{3}\) by 7 = 1
So, Answer will be E
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.
=> \(333^{222}\) = \((329 + 4)^{222}\)
Watch this video to MASTER BINOMIAL Theorem
Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 222C222 * 329^0 * 4^222) is divided by 7
=> Remainder of 4^222 is divided by 7
=> Remainder of 2^444 divided by 7
To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7
Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1
=> Cycle is 3
Remainder of 444 by 3 = Remainder of (4+4+4) by 3 = 0
=> Remainder of \(2^{444}\) by 7 = Remainder of \(2^{Cycle}\) by 7 = Remainder of \(2^{3}\) by 7 = 1
So, Answer will be E
Hope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem
28 Dec 2024, 23:49
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Which other numbers are like 7? For which numbers should we use Cyclic reminder approach and which numbers we should we Binomial approach. Please explain.
Bunuel
29 Dec 2024, 00:19
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Jyoti1812
Check LAST DIGIT OF A POWER chapter here: https://gmatclub.com/forum/math-number- ... 88376.html
