Powers and roots: Someone sort this out pls...

[uzoik]

Been looking for this question on the forum but cant find it. Can someone kindly help?

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

A) 1
B) 2
C) 3
D) 4
E) 5

Cheers!

[netigen]

if the OA is 4 then this is the soln (I didn't use paper for this so plz pardon my shortcuts)

The remainders for eq 3^X/5 will repeat at an interval of 4

3^1/5 = 3
3^2/5 = 4
3^3/5 = 2
3^4/5 = 1
3^5/5 = 3
3^6/5 = 4
3^7/5 = 2
3^8/5 = 1

8n+3 is of the form 4 x (2n) + 3

so the remainder for 3^(8n+3) will be 2

so the remainder for 3^(8n+3)+2/5 will be = 4

[lunarpower]

the above post, featuring pattern recognition, is great.

--

you can also try plugging in. here are two ways.

(1) just go with it

pick the smallest positive integer, namely 1. then you want the remainder when 3^11 + 2 is divided by 5. that's a fairly horrible number, but it's nothing you can't compute within two minutes (as long as you get started on it right away). here's a shortcut:

3^11 = (3^5)(3^5)(3)
= 243 x 243 x 3
= 177,147

plus two = 177,149
this ends with a 9, so the remainder is 4.

of course, this calculation is completely unnecessary; all you have to do is realize that the units (ones) digit is all that matters, since remainders upon division by 5 depend only on that digit. so, instead of figuring the actual value of 243 x 243 x 3, all you have to consider is
...3 x ...3 x 3 = ...7 (because 3 x 3 x 3 = 27). adding two gives ...9, so the remainder is 4.

(2) plug in 0
yes, i know that 0 is not a positive integer. however, plugging in 3, 2, and 1 give the following numbers:
3^27 + 2
3^19 + 2
3^11 + 2
plugging in 0 gives 3^3 + 2. since this continues to follow the same pattern evident in the above numbers, there's no reason to assume the pattern won't continue.

sure enough, plugging in 0 gives 3^3 + 2 = 29, whose remainder upon dividing by 5 is 4.

they're only saying 'positive' to scare you away from plugging in zero; they're onto you that way. (on the other hand, you actually can't plug in negative numbers, because those will be decimals.)

[resilient]

Ron,

I am always attracted to picking numbers and it has become second nature for me. BUT, I am following your approach here and seeing that the remainder is 8. 177149/5 is 35429.8. Are we missing parentheses or something

[lunarpower]

Ron,

I am always attracted to picking numbers and it has become second nature for me. BUT, I am following your approach here and seeing that the remainder is 8. 177149/5 is 35429.8. Are we missing parentheses or something

nope, you're doing everything right. you just have to know how to interpret remainders when you see them as decimals.

here's an example: let's say you divide 17 by 4. you already know the answer to this: it's 4, with a remainder of 1. but if you do long division, you'll see 4.25. does this mean you're done something wrong? no, of course it doesn't: it means that the decimal part is the remainder DIVIDED BY THE NUMBER YOU'RE ORIGINALLY DIVIDING BY. so, the decimal part (.25) isn't 1; it's 1/4. if you multiplied back by 4 (.25 x 4), you'd get the actual remainder.

so, in this problem, there are 2 ways to realize that the remainder is actually 4:
(1) multiply back by 5: .8 x 5 = remainder = 4.
(2) think about decimal/fraction equivalents: .8 is 4/5.

--

in any case, all this discussion is somewhat superfluous, because you don't get a calculator on the test! if you're doing long division, just stop long division at the point when you have a remainder. if you actually divide this out by hand all the way to the .8 part, then you're doing too much work; when you get the integer quotient with the 4 on the bottom, that's your remainder. end of story.

[II]

Hi Ron ... or anybody else.

How can you spot the trap answers for this question ?

For Your Information: My approach to answer this question.

Things to note: divisibility rule for the integer "5" is that if the last digit of the integer is 5 or 0, then the integer is divisible by 5. So we are only interested in the last digit.

First simplify the info in the question using the exponent rules:

3^(8n+3) => ((3^8 )^n)(3^3) => [(3^4)(3^4)]^n (27) =>
[(81)(81)]^n (27)

since we are only interested in the last digit we can multiply the last digits of 81, 81, and 27 --> 1*1*7 = 7.

7 + 2 = 9.

9/5 will leave remainder of 4.

[sudhir3127]

Hi the answer as discussed is 4.

My approach :

let n=1 the equation will be 3^11+2

we need to focus on units place and know that in case of 3 .. it repeats after every 4. hence the units place is 7 .

7+2 = 9 /5 . there fore the remainder is 4.

[lunarpower]

Hi the answer as discussed is 4.

My approach :

let n=1 the equation will be 3^11+2

we need to focus on units place and know that in case of 3 .. it repeats after every 4. hence the units place is 7 .

7+2 = 9 /5 . there fore the remainder is 4.

wow, that's brilliant.
i now feel utterly stupid for paying attention to more than just the units digit in my post above.

yes, yes, everyone look here - this is the most efficient solution. since this is division by 5, you only have to look at the units digit. i now owe this poster a beer, if he lives close enough to the san jose area.

note that if the division were by any number other than 2, 5, or 10, you couldn't just look at the units digit (and for divisibility by some numbers, such as 3, 9, and 7, you'd actually need the whole number.)

[lunarpower]

How can you spot the trap answers for this question ?

i'm not sure whether "trap answer" is even a meaningful concept on this problem, given that the answer choices comprise all possible remainders upon division by 5: 0, 1, 2, 3, and 4.

i mean, if your answer choices contain every answer that's even remotely possible, then can you still say that any of them are trap answers?