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If n is an integer greater than 0, what is the remainder when 9^(12n +

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KSBGC
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If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   Updated on: 26 Apr 2018, 06:12
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If n is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9
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Official Answer
E

Originally posted by KSBGC on 26 Apr 2018, 05:59.
Last edited by Bunuel on 26 Apr 2018, 06:12, edited 1 time in total.
Edited the question.
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   26 Apr 2018, 06:09
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selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n}+3\) is divided by 10?

a) 0
b) 1
C) 2
D)7
E)9



Can you please confirm the answer? I think remainder should be 4 that is not there in the options
3^24n can be 3^24 or any multiple of 24 i.e 24, 48 or 72
Now, this will have unit's digit =1(since 3^4=81 and 3 has cyclicity of 4)
Therefore units digit of the equation is 4 and remainder when divided by 10 is 4
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   26 Apr 2018, 06:09
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selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n}+3\) is divided by 10?

a) 0
b) 1
C) 2
D)7
E)9


If the question asks for the remainder when \(9^{12n+3}\), then the answer is 9 (E).

If the question asks for the remainder when \(9^{12n}+3\), then the answer must be 4 (not given among answer choices).

Please correct this minor error!
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   26 Apr 2018, 06:13
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Tulkin987 wrote:
selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n}+3\) is divided by 10?

a) 0
b) 1
C) 2
D)7
E)9


If the question asks for the remainder when \(9^{12n+3}\), then the answer is 9 (E).

If the question asks for the remainder when \(9^{12n}+3\), then the answer must be 4 (not given among answer choices).

Please correct this minor error!


Yes, both of you are right. Edited the question.
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   26 Apr 2018, 06:33
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selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9


\(9^{12n+3}\)- whatsoever value n takes(both even or odd), the final product of 12*n would always be even.
Even(12n) + Odd(3) = Odd

Now, if \(9^Even\) then the last digit(after expanding) is always 1
&, if \(9^Odd\) then the last digit(after expanding) is always 9
Upon division by 10 of expansion of \(9^Odd\) the remainder would be 9(Option E is the answer)
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If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   26 Apr 2018, 06:47
selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9



To solve this problem properly, we need to know the cyclicity of 9.
Every odd multiple always ends with a 9 and even multiple ends with a 1

12n+3 is odd for all value of n because 12n is always EVEN and EVEN + 3 = ODD

Therefore, the remainder when \(9^{12n+3}\) is divided by 10 is 9(Option E)
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   27 Apr 2018, 09:52
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selim wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9


For any allowable value of n, we see that the exponent (12n + 3) will be an odd number.
Let’s consider the pattern of units digits for 9 raised to positive integer powers. We see that 9^1 = 9, 9^2 = 1, 9^3 = 9, 9^4 = 1.

The pattern of units digits indicates that when 9 is raised to an odd exponent, the units digit is 9.

We may also recall that to determine the remainder when a number is divided by 10, we only need to divide the units digit by 10. Since 9/10 has a remainder of 9, we have our answer.

Answer: E
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink] New post   24 Apr 2023, 22:59
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KSBGC wrote:
If n is an integer greater than 0, what is the remainder when \(9^{12n+3}\) is divided by 10?

A. 0
B. 1
C. 2
D. 7
E. 9

Solution:

  • n is a positive integer
  • We are asked the value of \((\frac{9^{12n+3}}{10})_R\)
  • we know that a number when divided by 10 will give the remainder same as its unit digit
  • So, \((\frac{9^{12n+3}}{10})_R=(9^{12n+3})_U\)

  • Cyclicity of 9 is 2
  • So, \((9^{even})_U=1\) and \((9^{odd})_U=9\)
  • We see that \(12n+3=odd\) and therefore \((\frac{9^{12n+3}}{10})_R=(9^{12n+3})_U=(9^{odd})_U=9\)

Hence the right answer is Option E
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Re: If n is an integer greater than 0, what is the remainder when 9^(12n + [#permalink]
24 Apr 2023, 22:59
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