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Re: What is the remainder when (92n + 3)(5n ) is divided by 10, where n is
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22 Sep 2024, 17:11
1
The cycle of unit digits of 9 is 9, 1 (9^1 = 9, 9^2 = 81, 9^3 = 729 and so on) The cycle of unit digits of 9 is 5 (5^1 = 5, 5^2 = 25..) Multiply cycle of unit digits of 9 and 5 together, 9*5 = 45, 1*5 = 5 We see that unit digits is always 5, so remainder is 5 when the number is divided by 10
Re: What is the remainder when (92n + 3)(5n ) is divided by 10, where n is
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24 Sep 2024, 11:50
1
The remainder when a number is divided by 10 is always the unit digit. Now when you raise 5 to any power of positive integer, unit digit is 5. 5 multiplied by any number leaves a remainder of either 0 or 5 as unit digit. Since there is no zero in the option, answer is 5.
Re: What is the remainder when (92n + 3)(5n ) is divided by 10, where n is
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16 Jan 2025, 03:59
1
What is the remainder when \((9^{2n + 3})(5^n)\) is divided by 10
Theory: Remainder of a number by 10 is same as the unit's digit of the number
(Watch this Video to Learn How to find Remainders of Numbers by 10)
Using Above theory Remainder of \((9^{2n + 3})(5^n)\) by 10 = Unit's digit of \((9^{2n + 3})(5^n)\)
Now, Let's find the unit's digit of \((9^{2n + 3})\)
Now to find the unit's digit of \(9^{2n + 3}\), we need to find the pattern / cycle of unit's digit of power of 9 and then generalizing it. Unit's digit of \(9^1\) = 9 Unit's digit of \(9^2\) = 1 Unit's digit of \(9^3\) = 9 Unit's digit of \(9^4\) = 1 => Units' digit of Odd Power of 9 = 9 => Units' digit of Even power of 9 = 1
2n + 3 = Even + Odd = Odd => Unit's digit of \(9^{2n + 3}\) = 9
Unit's digit of \(5^n\) = 5 (as all positive powers of 5 have units' digit of 5)
=> Unit's digit of \((9^{2n + 3})(5^n)\) = Unit's digit of 9 * 5 = 5
So, Answer will be D Hope it helps!
MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem
gmatclubot
Re: What is the remainder when (92n + 3)(5n ) is divided by 10, where n is [#permalink]