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If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A 0 B 1 C 2 D 3 E 4

It should be \(3^{8n+3}+2\).

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

ahhh okay. i followed the same pattern for the exponents. the problem made it seem like +2 was in the exponent. makes perfect sense now. much appreciated, Bunuel.
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put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???

counter arguments welcome instead of kudos !!!

n is given to be a positive integer so it cannot be 0. That said, whatever will be the last digit of \(3^{(8n+3)} + 2\) for n = positive integer, will be the last digit when n = 0. (We know cyclicity works for n = 0 too) Hence your short cut works.
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Answer should be E i.e. 4 remember 3 has a cyclicity of 4 =>3,9,*7,*1 3^(8n+3) +2 => for n=1 => 8*1+3 =>11 =>3(on dividing 11/4) for n=2 =>19=>3(on dividing 19/4) for n=3 =>27=>3(on dividing 27/4) for n=4 =>35=>3(on dividing 35/4) for n=5 =>43=>3(on dividing 43/4) for n=6 =>51=>3(on dividing 51/4) therefore 3^3 gives 7 in units place => 7+2 => 9 in all the cases hence on dividing 9/5 remainder is 4
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Re: If n is a positive integer, what is the remainder when [#permalink]

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06 Jun 2013, 14:00

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you can split it up to make it look simpler. if x = 1 (3^8)(3^3) + 2 3^8 = 9^4 = 9 x 9 x 9 x 9 = 81 x 81 (units digit is 1) 3^3 = 27 (units digit is 7) 1 x 7 + 2 = 9 9/5 = 1 r4 Answer is E

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A 0 B 1 C 2 D 3 E 4

It should be \(3^{8n+3}+2\).

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A 0 B 1 C 2 D 3 E 4

It should be \(3^{8n+3}+2\).

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question

The reason we are focusing on \(3^3\) is because the repeating cycle of \(3^{8n}\) only matters because it is a sequence that loops in a pattern of 4 (which is consequently a divisor of 8).

You said that you tested the cycle. So you should understand that, as Bunuel said, 3-9-7-1-3-9-7-1 is the cycle. No matter what multiple of 8 you use (8, 16, 24, 32, etc.), 1 will always be the units digit of the final value. If that is the case we can actually forget the 8n part of the equation, because you know that \(3^{8n}\) will always have a units digit of 1. This means that you have to focus on the \(3^3\) instead. You can look at this is two ways. Either go three more steps in the cycle and land on 7 (3-9-7), or because \(3^{8n+3} = 3^{8n}*3^3\) = (number with the units digit of 1) x 27. The units digit will be 7 because 1 x 7 is 7. 7+2 = 9, 9/5 = 1 r4. Answer is E.

You can also look at my post above for a method that doesn't involve all this sequencing stuff

On another note, be careful not to make this the sequencing a rule for all questions that look like this because it definitely is not a rule. If you see \(3^{7m+3}\), you can't just "cancel out" the 7m. If m=1 the cycle is 3-9-7-1-3-9-7, but if m=2 it continues the sequence and does not restart back at 3 but instead starts where it left off after 7 (1-3-9-7-1-3-9).

The reason we focus on the \(3^3\) in this equation is only because the final value of the "8n sequence" always ends with the same units digit.

Re: If n is a positive integer, what is the remainder when [#permalink]

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07 Jun 2013, 01:11

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azule45 wrote:

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0 B. 1 C. 2 D. 3 E. 4

We know that Remainder(\(\frac{a*b}{c}\)) = Remainder(\(\frac{a}{c}\))*Remainder(\(\frac{b}{c}\)) for integral values of a,b and c.

The given expression : \(\frac{(3^{8n}*3^3 + 2)}{5}\) = \(\frac{27*(3^4)^{2n}}{5}\)+\(\frac{2}{5}\) = Remainder of \(\frac{(27)*(81)^{2n}}{5}\) + Remainder of \(\frac{2}{5}\) = Remainder of\(\frac{(2)*(1)^{2n}}{5}\)+Remainder of \(\frac{2}{5}\)= Remainder of \(\frac{4}{5}\) = 4.
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Re: If n is a positive integer, what is the remainder when [#permalink]

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03 Feb 2014, 09:32

azule45 wrote:

If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0 B. 1 C. 2 D. 3 E. 4

I followed a slightly longer approach

Assume n=1 then 3^11 + 2 /5, what is the remainder?

Well 3^11 is (5-2)^11 so we have -2^11/5 what is the remainder of this?

Again -2*4^5 = -2*(5-1)^5, which gives a remainder of -1 multiplied by -2 gives a remainder of 2, and if we add the 2 that was missing for the first part of the question we get 4 as a remainder

Re: If n is a positive integer, what is the remainder when [#permalink]

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09 Apr 2014, 07:03

\(\frac{(3^8)^n*3^3 + 2}{5}\)

\(\frac{81^n * 81^n * 27 + 2}{5}\) Making use of binomial theorem:

\(\frac{ (80+1)^n * (80+1)^n * 27 + 2 }{5}\)

\(\frac{ 1^n * 1^n * 2 + 2 }{5}\)

\(\frac{2+2}{5}\)

\(\frac{4}{5}\)

Thus 4 is remainder.
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Look at the last digits of powers: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9...

So whenever we have a power of 3, we remove groups of 4 out of it i.e. divide it by 4 and whatever is the remainder tells us the last digit.

So say we have 3^13, we divide 13 by 4 and are left with 1 remainder. So this is the first term of the next cycle and will end with 3. Say we have 3^39, we divide 39 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.

If you have \(3^{8n+3}\), we divide 8n+3 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7.

Last term of \(3^{8n+3}+2\) will be 7+2 = 9. When you divide a number that ends with 9 by 5, you get 4 remainder. If you are not sure why this is so, check out the first part of the question discussed in this post: http://www.veritasprep.com/blog/2014/03 ... emainders/

Answer (E)

Or since you know that cyclicity of 3 is 4, just put n = 0 to get 3^3 + 2 = 29. When you divide 29 by 5, you get remainder 4.
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