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Remainder when divided by 5

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Re: Remainder when divided by 5

by Vemuri » Tue May 05, 2009 9:46 am
(3^8n+3) + 2 ==> 3^(8n)+5 (I am assuming this is what the question is saying). If this is correct, then the reminder should be 1 (because 3^(8n) will always have its unit digit as 1. If 5 is added to the number, then dividing the number by 5 should leave a reminder of 1.

Please check the question
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by sureshbala » Tue May 05, 2009 10:24 pm
We need the remainder of 3^[(8n+3)]+2 when divided by 5.

Let us calculate the remainder of 3^(8n+3)

3^(8n+3)
= 3^8n x 3^3
= (3^4)^2n x 27

Since 3^4 =81 leaves a remainder 1 when divided by 5, the remainder of 3^8n when divided by 5 is always 1.

Also 27 leaves remainder 2 when divided by 5.

Hence the remainder of 3^(8n) x 27 i.e 3^(8n+3) when divided by 5 is 2.

Thus the remainder of 3^[(8n+3)]+2 = 4.

Of course you can also conclude this by taking n=1 since there is no option like Cannot be determined or none of these
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by crackgmat007 » Wed May 06, 2009 10:06 am
Hi Bala,

Is this correct?

Find out the units digit.

3^(8n+3)
= 3^8n x 3^3
= (3^4)^2n x 27
units digit for (3^4)^2n is '1'; units digit of 27 is '7'. Hence the units digit of the product is 7, when divided by 5 leaves a remainder of 2. Hence 2+2 = 4.
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by sureshbala » Wed May 06, 2009 10:49 am
crackgmat007 wrote:Hi Bala,

Is this correct?

Find out the units digit.

3^(8n+3)
= 3^8n x 3^3
= (3^4)^2n x 27
units digit for (3^4)^2n is '1'; units digit of 27 is '7'. Hence the units digit of the product is 7, when divided by 5 leaves a remainder of 2. Hence 2+2 = 4.
Yes, since the remainder obtained when a number is divided by 5 or 10 is the units of a number this can always be done...
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