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02 Apr 2020, 05:06
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Find the remainder when \(25^{18}\) is divided by 9.
A. 1
B. 2
C. 3
D. 5
E. 6
A. 1
B. 2
C. 3
D. 5
E. 6
02 Apr 2020, 05:13
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Dillesh4096
\(R[\frac{25^{18}}{9}] = R[\frac{(27-2)^{18}}{9}] = R[\frac{(-2)^{18}}{9}] = R[\frac{(2)^{18}}{9}] \)
\(R[\frac{(2)^{3}}{9}] = 8 or -1\) (negative remainder)
CONCEPT: Remainder can be written in both +ve and -ve form for simplification. When divisor is 9 and remainder is 7 then the number has either 7 in excess or number is 2 short to be divisible by 9 hence
Remainder (+7) = Remainder (-2) when divisor is 9
Remainder (+7) = Remainder (-2) when divisor is 9
Taking exponent 6 both sides
\(R[\frac{(2)^{18}}{9}] = (-1)^6 = 1 \)
Answer: Option A
The concept video of remainder theorem is as follows:
27 Sep 2020, 08:57
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A method which can be used in such a case is using cyclicity of remainders. In this we see how many cycles are required to get a remainder of 1.
Interestingly, if you get a remainder of -1, then the double of the cycles gives a remainder of 1.
We get a remainder of -1, if the remainder is 1 less than the divisor (eg, 31 divided by 8 gives a remainder of 7 which is also -1)
We stop the cycle, when we get a remainder of 1 or 8. If we get a remainder of 8, then double that cycle will give a remainder of 1.
\(R[\frac{(25)^{18}}{9}] = R[\frac{5^{36}}{9}]\)
Cycle 1: \(R[\frac{5}{9}] = 5\)
Cycle 2: We multiply the remainder of the previous cycle by 5 and get the next remainder. \(R[\frac{5 * 5}{9}] = R[\frac{25}{9}] = 7\)
Cycle 3: \(R[\frac{5 * 7}{9}] = R[\frac{35}{9}] = 8 \space or -1\)
Therefore, we shall get a remainder of 1 in the 6th cycle. This means that the remainder for every 6 powers of 5 divided by 9 = 1
\(R[\frac{5^6}{9}] = 1\)
Therefore \(R[\frac{5^{36}}{9}] = R[\frac{(5^6)^6}{9}] = R[\frac{1}{9}] = 1\)
Option A
Arun Kumar
Interestingly, if you get a remainder of -1, then the double of the cycles gives a remainder of 1.
We get a remainder of -1, if the remainder is 1 less than the divisor (eg, 31 divided by 8 gives a remainder of 7 which is also -1)
We stop the cycle, when we get a remainder of 1 or 8. If we get a remainder of 8, then double that cycle will give a remainder of 1.
\(R[\frac{(25)^{18}}{9}] = R[\frac{5^{36}}{9}]\)
Cycle 1: \(R[\frac{5}{9}] = 5\)
Cycle 2: We multiply the remainder of the previous cycle by 5 and get the next remainder. \(R[\frac{5 * 5}{9}] = R[\frac{25}{9}] = 7\)
Cycle 3: \(R[\frac{5 * 7}{9}] = R[\frac{35}{9}] = 8 \space or -1\)
Therefore, we shall get a remainder of 1 in the 6th cycle. This means that the remainder for every 6 powers of 5 divided by 9 = 1
\(R[\frac{5^6}{9}] = 1\)
Therefore \(R[\frac{5^{36}}{9}] = R[\frac{(5^6)^6}{9}] = R[\frac{1}{9}] = 1\)
Option A
Arun Kumar
