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Remainder Problem

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by jnellaz » Fri Nov 07, 2008 9:46 am
I came across this problem but I am trying to figure out how they got to the answer. Can someone provide their approach to this problem? Thanks!


If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4

[spoiler]Answer: E or 4[/spoiler][spoiler]
[/spoiler]
Last edited by jnellaz on Fri Nov 07, 2008 10:08 am, edited 2 times in total.
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Re: Remainder Problem

by logitech » Fri Nov 07, 2008 9:59 am
jnellaz wrote:I came across this problem but I am trying to figure out how they got to the answer. Can someone provide their approach to this problem? Thanks!


If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4

[spoiler]Answer: 4[/spoiler][spoiler]
[/spoiler]
Two things:

1) There seems to be a typo
2) Does answer 4 mean 4 or D ? :)
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by jnellaz » Fri Nov 07, 2008 10:11 am
Hey there,

I fixed the spoiler function to show the correct answer choice - but the answer is 4 or choice E.

Where do feel there is a typo? I got this question from here:
https://www.urch.com/forums/gmat-problem ... -help.html
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by logitech » Fri Nov 07, 2008 10:38 am
jnellaz wrote:Hey there,

I fixed the spoiler function to show the correct answer choice - but the answer is 4 or choice E.

Where do feel there is a typo? I got this question from here:
https://www.urch.com/forums/gmat-problem ... -help.html
38n+3 + 2 is divided by 5?

It is not really CONCISE to write the question this way

38n+5 is more clear, and the remainder changes with different n values. So there is a typo

but if it were written as this:

38^(n+3) + 2

we will still get different remainders..

last option would be:

3^(8n+3)+2

AHA :idea:

3^(8n+3) will always have 7 at the unit digits

and we add 2 so the unit digit will always be 9

and divided by 5...

will ALWAYS give us 4

So bottom line is: TYPOS are waste of time, but we still can learn something while figuring out what they are! :D
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by jnellaz » Fri Nov 07, 2008 10:53 am
Thanks for your post! How did you quickly figure out that 3^(8n+3) will always have 7 at the unit digits?
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by logitech » Fri Nov 07, 2008 10:58 am
jnellaz wrote:Thanks for your post! How did you quickly figure out that 3^(8n+3) will always have 7 at the unit digits?
the unit digit of power of three are 3,9,7, and 1 than they all repeat each other. Just play with the numbers and you will find out.
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by jimmiejaz » Fri Nov 07, 2008 11:01 am
jnellaz wrote:Thanks for your post! How did you quickly figure out that 3^(8n+3) will always have 7 at the unit digits?
Let me take this ahead logitech!!!!!
if you check 3 powers have the unit digit 3 after every 4 powers...
ie. 3^1=3 , 3^5=243
3^2=9, 3^6=729
3^3=27,3^7=2187
so, i hope u get the pattern now....
So put n =1, we get 3^11, n=2 we get 3^19 so.. the units digit will be the same as 3^3 or 7.
hope it helps....
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by jnellaz » Fri Nov 07, 2008 11:13 am
Got it. Thanks to you both!

And for those of you out there that have read this and want a quick list of repeatable units for 2-12 powers here it is:

Some examples:

2^5 = 32 = Unit is 2.
5^12 = 244140625 =Unit is 5

2 = 2, 4, 8, 6
3 = 3, 9, 7, 1
4 = 4, 6
5 = 5
6 = 6
7 = 7, 9, 3, 1
8 = 8, 4, 2, 6
9 = 9, 1
10 = 0
11 = 1
12 = 2, 4,8, 6
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by singalong » Sat Nov 19, 2011 1:22 am
logitech wrote:
jnellaz wrote:Thanks for your post! How did you quickly figure out that 3^(8n+3) will always have 7 at the unit digits?
the unit digit of power of three are 3,9,7, and 1 than they all repeat each other. Just play with the numbers and you will find out.
I am not able to connect the dots here.I understood that unit digit of power of three are 3, 9, 7 and 1.They form a cycle of 4.But how is 3^(8n+3) gonna have 7 at it's unit digit?
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hi

by Scott@TargetTestPrep » Wed Dec 20, 2017 7:34 am
If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?

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

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