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

Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1.
shouldn't it be 3^0=1 instead?

No, as in this case all numbers in power which is multiple of cyclicity would have the last digit of 1.

When power is divisible by cyclicity # (remainder 0), then the last digit is the same as the last digit of number in the power of cyclicity #.

For example:
Cyclicity of 2 is 4, so \(2^{4n}\) (where \(n\) is a positive integer) will have the same last digit as \(2^4\), which is 6.

OR:

Cyclicity of 4 is 2, so \(4^{2n}\) (where \(n\) is a positive integer) will have the same last digit as \(4^2\), which is 6.

Hope it's clear.

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.

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

5 is a special case because every number with a unit's digit of 5 or 0 is divisible by 5. So whenever you have a number that ends in 9, the number that is 4 less than it which ends in 5 will be a multiple of 5.
Say you get 33759, then definitely 33755 will be a multiple of 5 so your remainder will be 4.
Similarly a number ending with 7 will always give 2 as remainder when divided by 5; a number ending in 3 will give 3 as remainder when divided by 5 and so on.
You can say the same thing only about one more number i.e. 2. Every time the unit's digits is 1/3/5/7/9 remainder will be 1. Every time unit's digits is 0/2/4/6/8 the remainder will be 0.

If you really want to get to the concept, think about this:
Is it just co-incidence that unit's digits have a cyclicity of 10 (i.e. after 10 numbers you get the same unit's digits) and you can find the remainder using just the unit's digit for only 2 and 5, the factors of 10?

Note: Getting into concepts and linking one with the other to get the full picture is what leads to a 50-51 in Quant.

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

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.

Kudos if that helps.

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.

Let us plug in:

n = 1; 3^(11) + 2

Use cylicity:
Units digit:
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
and so on.
3^11 will have 7 at the end hence the remainder will be 7 + 2 = 9/5 = +4

You should know the concept of cyclicity. Powers of 3 have a cyclicity of 4 i.e.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.

We need to determine the remainder of:

3^8n x 3^3 + 2 when it is divided by 5.

We only need to know the units digit of the above expression to determine the remainder.

Let’s look at the pattern of units digits of powers of 3. Note that we are only concerned with the units digits, so, for example, for 3^3, we concern ourselves only with the units digit of 27, which is 7. Here is the pattern:

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The repeating pattern is 3-9-7-1. Since the units digit pattern for a base of 3 is 3-9-7-1, we see that whenever 3 is raised to an exponent that is a multiple of 4, the units digit will be 1. Thus:

3^8n x 3^3 = units digit of 1 x units digit of 7 = units digit of 7

So, units digit of 7 + 2 = units digit of 9, and thus 9/5 has a remainder of 4.

Answer: E

Hi VeritasKarishma - i was able to understand what you said in blue completely

But not sure what are you referring to in the pink font above ?

could you please share link for the purple bit specifically

This is the concept of cyclicity in remainders. Please check it from a GMAT specific resource.

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