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# If a and b are positive integers, what is the remainder when 9^2a+b

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Joined: 17 Jan 2017
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If a and b are positive integers, what is the remainder when 9^2a+b  [#permalink]

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05 Apr 2018, 06:32
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62% (00:45) correct 38% (01:21) wrong based on 58 sessions

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If a and b are positive integers, what is the remainder when $$9^{2a+b}$$ is divided by 20?

(1) a = 3
(2) b = 5

Source: ExpertsGlobal
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Joined: 25 Feb 2013
Posts: 1185
Location: India
GPA: 3.82
Re: If a and b are positive integers, what is the remainder when 9^2a+b  [#permalink]

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05 Apr 2018, 11:04
1
benejo wrote:
if a and b are positive integers, what is the remainder when $$9^{2a+b}$$ is divided by 20?

(1) a = 3
(2) b = 5

Source: ExpertsGlobal

$$9^{2a+b}$$ to know the remainder of this number when it is divided by 20 we need to know the last two digits of this number, i.e. we need to know the units and tens digit of this number.

The cyclicity of $$9$$ is as follows -

$$9^1=9; 9^2=81; 9^3=729; 9^4=6561; 9^5=59049$$

So units digit of the number will be either $$1$$ or $$9$$ and tens digit will always be an even number. Hence the remainder of $$9^{2a+b}$$ when divided by $$20$$ will depend on the UNITS digit of this number.

Now $$2a$$ is always EVEN irrespective of the value of $$a$$. Hence $$9^{{Even}+b}$$. So we only need to know the value of $$b$$ to get the units digit.

Statement 1: Nothing mentioned about $$b$$. Insufficient

Statement 2: $$b$$ is odd so $$9^{2a+b}=9^{{Even}+Odd}=9^{Odd}$$. Hence the remainder will be $$9$$. Sufficient

Option B
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Joined: 02 Sep 2009
Posts: 47219
Re: If a and b are positive integers, what is the remainder when 9^2a+b  [#permalink]

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05 Apr 2018, 11:50
benejo wrote:
if a and b are positive integers, what is the remainder when $$9^{2a+b}$$ is divided by 20?

(1) a = 3
(2) b = 5

Source: ExpertsGlobal

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Re: If a and b are positive integers, what is the remainder when 9^2a+b  [#permalink]

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12 Apr 2018, 22:13
Please explain to me why we can make sure that 9^odd (I know that the unit digit of which is always 9) when divided by 20 doesn't leave a remainder of 19 instead of 9? Thanks!
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Joined: 02 Sep 2009
Posts: 47219
Re: If a and b are positive integers, what is the remainder when 9^2a+b  [#permalink]

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13 Apr 2018, 00:00
1
vap311096 wrote:
If a and b are positive integers, what is the remainder when $$9^{2a+b}$$ is divided by 20?

(1) a = 3
(2) b = 5

Please explain to me why we can make sure that 9^odd (I know that the unit digit of which is always 9) when divided by 20 doesn't leave a remainder of 19 instead of 9? Thanks!

When x9 is divided by 20, the remainder could be either 9 or 19.

If x is odd, then x9 is 19, 39, 59, ..., all of which give the remainder of 19 upon division by 20.
If x is even, then x9 is 09, 29, 49, ..., all of which give the remainder of 9 upon division by 20.

Now, 9^odd, not only results in the units digit of 9 but also in the even tens digit: 9^1 = 09, 9^3 = 729, 9^5 = 59,049, ... Thus, 9^odd = ...(even)(9), which as we saw above will always give the remainder of 9 upon division by 20.

Hope it's clear.
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Re: If a and b are positive integers, what is the remainder when 9^2a+b &nbs [#permalink] 13 Apr 2018, 00:00
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