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Re: What is the remainder when 43^{86} is divided by 5? [#permalink]

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21 Jun 2012, 17:58

9

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Alterego wrote:

What is the remainder when \(43^{86}\) is divided by 5?

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

Please provide a detail explanation on how you achieved the correct answer. Thanks

First, you have to come into terms that the GMAT doesn't expect you to calculate for the value of 43^86.

Second, you have to know that when it comes to these kinds of questions, the only digit that matters is the units digit of the number.

Always try to enumerate the powers of the said number to LOOK FOR THE PATTERN:

3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 (see it's still easy to multiply 3 from the previous digit, it's still "time-friendly") 3^5 = 243 (it's still time-friendly here) 3^6 = (now it becomes counter-productive to calculate 243*3; so what do we do then? let's just multiply the units digit by 3) = 3*3 = 9 3^7 = _ _ _ 7 (7 is the last digits, although I don't know if it's a four digit number of 5, doesn't matter)

Are you seeing the pattern? If you haven't, check out the corresponding units digit for each power

when raised to 1, the units digit is 3 raised to 2, the units digit is 9 raised to 3, the units digit is 7 raised to 4, the units digit is 1 raised to 5, the units digit is 3 <--- "the cycle begins again" raised to 6, the units digit is 9

Now we know that raised to 6, the units digit is 9, the question says that 43 should be raised to 86 (which is equal to raised to 6, check our pattern). This means the units digit is 9

Now let's divide 9 by 5

What's the remainder? 4

Answer: (E)

(kudos? )
_________________

Far better is it to dare mighty things, to win glorious triumphs, even though checkered by failure... than to rank with those poor spirits who neither enjoy nor suffer much, because they live in a gray twilight that knows not victory nor defeat. - T. Roosevelt

Since 40 is completely divisible by 5, you only have to think about \(3^{86}\) \(3^{86} = 9^{43} = (10 - 1)^{43}\)

Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be \((-1)^{43} = -1\) which means the remainder is 5 - 1 = 4

What is the remainder when 43^86 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4

Notice that \(43^{86}=(40+3)^{86}\). Now, if we expand this expression, all terms but the last one will have 40 as multiple and thus will be divisible by 5. The last term will be \(3^{86}\). So we should find the remainder when \(3^{86}\) is divided by 5.

Next, \(3^{86}=9^{43}\). 9 in odd power has units digit of 9 hence yields the remainder of 4 upon division by 5 (9 in even power has units digit of 1 hence yields the remainder of 1 upon division by 5).

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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22 Jun 2012, 13:22

i did it the following way (just making sure that its not by luck that i got the right answer)

43^86 /5 ==> since we are looking for a remainder that is 40+3 ... 3^86 [3 repeats in the following manner 3^1=3, 3^2=9, 3^3=7, 3^4=1 [only unit digits]

86/4 ==> Remainder is 2 .. which means that unit digit is going to be 9 (9/5 ==> gives a remainder of 4 as E)

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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22 Jun 2012, 16:56

Thanks for all your replies guys. It was definitely helpful. Veritasprepkaris, those links were very useful, thanks. Bunuel, those similar practice questions was a great idea, thanks.

i did it the following way (just making sure that its not by luck that i got the right answer)

43^86 /5 ==> since we are looking for a remainder that is 40+3 ... 3^86 [3 repeats in the following manner 3^1=3, 3^2=9, 3^3=7, 3^4=1 [only unit digits]

86/4 ==> Remainder is 2 .. which means that unit digit is going to be 9 (9/5 ==> gives a remainder of 4 as E)

i did it the following way (just making sure that its not by luck that i got the right answer)

43^86 /5 ==> since we are looking for a remainder that is 40+3 ... 3^86 [3 repeats in the following manner 3^1=3, 3^2=9, 3^3=7, 3^4=1 [only unit digits]

86/4 ==> Remainder is 2 .. which means that unit digit is going to be 9 (9/5 ==> gives a remainder of 4 as E)

Your logic worked because we are discussing divisibility by 5 here. The last digit decides the remainder when a number is divided by 5. Remainder when ****7 is divided by 5 will always be 2. Remainder when *****4 is divided by 5 will always be 4. This is so because every number that ends in 0 or 5 is divisible by 5 and only numbers ending in 0 or 5 are divisible by 5. Last digit works only for 2 and 5.

If you consider divisibility by say 3 or 7 etc last digit logic doesn't work so be careful.
_________________

Since 40 is completely divisible by 5, you only have to think about \(3^{86}\) \(3^{86} = 9^{43} = (10 - 1)^{43}\)

Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be \((-1)^{43} = -1\) which means the remainder is 5 - 1 = 4

Here 73^80 is number with unit's digit 1. There fore a number ending in 7 when raised to the power a number ending in 1 - the units digit will vary for example - unit's digit of xy7^11 = 3, unit's digit of xy7^21 = 7, unit's digit of xy7^31 = 3,

however the explanation is as shown in the file attached.

I didnt get the part "If the remainder is 1, that means we will take the first term of the cycle of 7^x.

Another similar problem : 28^(43^20)

here according to me units digit will vary according to the power:

unit's digit of xy8^11 = 2, unit's digit of xy8^21 = 8,

Since 40 is completely divisible by 5, you only have to think about \(3^{86}\) \(3^{86} = 9^{43} = (10 - 1)^{43}\)

Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be \((-1)^{43} = -1\) which means the remainder is 5 - 1 = 4

Here 73^80 is number with unit's digit 1. There fore a number ending in 7 when raised to the power a number ending in 1 - the units digit will vary for example - unit's digit of xy7^11 = 3, unit's digit of xy7^21 = 7, unit's digit of xy7^31 = 3,

however the explanation is as shown in the file attached.

I didnt get the part "If the remainder is 1, that means we will take the first term of the cycle of 7^x.

Another similar problem : 28^(43^20)

here according to me units digit will vary according to the power:

unit's digit of xy8^11 = 2, unit's digit of xy8^21 = 8,

Please guide.

This question uses two related but different concepts - cyclicity and remainders

What is the last digit of \(47^5\)?

You know that the last digit is obtained by focusing on the last digit on the number i.e. 7. 7 has a cyclicity of 7, 9, 3, 1

47^1 = 47 47^2 = ...9 47^3 = ....3 47^4 = ....1 47^5 = ....7 47^6 = ....9 47^7 = .....3 47^8 = ....1 and so on

So the last digit depends on the remainder obtained when the power is divided by 4. If the power is a multiple of 4, the last digit will always be 1. If the power is 1 more than a multiple of 4 (remainder 1 when power is divided by 4), the last digit will be 7. If the power is 2 more than a multiple of 4, the last digit will be 9. If the power is 3 more than a multiple of 4, the last digit will be 3.

The concept remains the same here: \(47^x\) where \(x = 73^{80}\) The last digit depends on whether x is divisible by 4 or not. If not, what is the remainder?

So basically we have to take two steps: Step I: Find the remainder when x is divided by 4. Step II: Find the corresponding last digit for that remainder.

Step I: \(73^{80}/4\) \((72+1)^{80}/4\) (This concept is discussed above) 72 is divisible by 4.

So when \(73^{80}\) is divided by 4, remainder is 1.

So the power of 47 is 1 more than a multiple of 4.

Step II Knowing the cyclicity of 7, this means the last digit will be 7.
_________________

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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11 Jun 2013, 02:12

Hi Guys, pretty new to the forum!

in such questions, one method which I follow is as follows:- for 43^86, we can find the last digit of this & that's enough for us.

consider the 3^86(units digit)

1) power cycle of 3 as explained by mates above is "4" 2) find out remainder of 86/4, which is "2" 3) take "2" as your new power 4) now find out remainder when (3^2) /5, which is 4

hope this helps! might not be your only method, but might be the best method!

Since 40 is completely divisible by 5, you only have to think about \(3^{86}\) \(3^{86} = 9^{43} = (10 - 1)^{43}\)

Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be \((-1)^{43} = -1\) which means the remainder is 5 - 1 = 4

ya it makes sense..i also did the same thing until the last step where 3^86, So rather than taking 9, i took 27(Since i wanted a positive remainder.). But this gives 2 as a remainder, although remainder should come same.
_________________

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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28 Jun 2013, 20:00

Alterego wrote:

What is the remainder when 43^86 is divided by 5?

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

An easy approach that will help you solve the question in less than a minute.

Now, we have 43^86 divided by 5

5*9= 45

If i divide, 43 by 5 i get the remainder as 3 or -2. I hope you get the -2 logic. ( 3-5 => -2)

remainder can be 3 or it can also be -2.

Now,using this approach. 43^86 can be written as (-2)^86

=> 43^86 => (-2)^86 => (-2)^84 * (-2)^2 ------ When the bases are same the powers get added, we can split the term -2^86 => (-2^4)^21 * (-2)^2 ----------- 84 can be rewritten as 4*21 and the 4 is brought inside the braces making -2 as -2^4 ---- remember this is all about playing with powers... so do it carefully....

=> 16^21 * 4

16^21 * 4 divided by 5 16 when divided by 5 gives remainder 1

1^21 *4 => 4 is the remainder...

You must be wondering when 86 was splitted to 84 and 2. See, we have to figure the power of 2 in such a way that it is easily divisible by 5. 2^1=2, when divided by 5 will give remainder 2...not of our use 2^2=4, when divided by 5 will give remainder 4...not of our use... this can be used to deduce the remained-- it is one of my favourite's apporach, but i will explain that later... 2^3=8, when divided by 5 will give remainder 3...not of our use 2^4=16, when divided by 5 will give remainder 1... this one is good.

Therefore, we get the remainder as 4..

Just practice few questions using this approach and it wont take more than a minute to solve remainder types of questions.... You can use the same approach for any number.
_________________

MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636 HEXAGON Theory ---> http://gmatclub.com/forum/hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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14 Jul 2013, 07:14

1

This post was BOOKMARKED

Here's how I did this one

\(\frac{3^{86}}{5}\)= ?

when

\(3^1\) divided by 5 remainder is 3 \(3^2\) divided by 5 remainder is 4 \(3^3\) divided by 5 remainder is (4*3) / 5 = 2 \(3^4\) divided by 5 remainder is ( 2*3)/ 5 = 1 \(3^5\) divided by 5 remainder is 3

So the repeating block is {3,4,2,1}

when we divide 86 by 4 the remainder is 2

Hence the remainder for this problem is 4. _________________

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Note: It is desirable to modify the expression in such a way that we get a remainder of 1, when divided by 5. Thus, we take out \(3^4\) from the given expression, which results in 81.

Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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28 Jul 2013, 04:27

Hope Everyone enjoys the shortest one:

Rem[ 43 ^ 86 / 5 ] ????

REM(43/5) = 3

=> REM( 3^86 /5) => REM( 9^43 /5) ?

Since REM(9/5) = 4

=> REM( 4^43/5)

=> REM ( 4^40 * 4 ^3)/5 ?

=> REM [ (16^10 * 16 *4)/5]

Since REM(16/5) = 1 , REM(4/5) = 4

=> REM(1*1*4)/5 = REM(4/5) = 4
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Re: What is the remainder when 43^86 is divided by 5? [#permalink]

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23 Sep 2013, 12:47

Bunuel wrote:

What is the remainder when 43^86 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4

Notice that \(43^{86}=(40+3)^{86}\). Now, if we expand this expression, all terms but the last one will have 40 as multiple and thus will be divisible by 5. The last term will be \(3^{86}\). So we should find the remainder when \(3^{86}\) is divided by 5.

Next, \(3^{86}=9^{43}\). 9 in odd power has units digit of 9 hence yields the remainder of 4 upon division by 5 (9 in even power has units digit of 1 hence yields the remainder of 1 upon division by 5).

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