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If you divide 7^131 by 5, which remainder do you get?

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If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post Updated on: 19 Oct 2012, 02:39
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If you divide 7^131 by 5, which remainder do you get?

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

Originally posted by blackcrow on 01 Sep 2009, 03:06.
Last edited by Bunuel on 19 Oct 2012, 02:39, edited 1 time in total.
Renamed the topic and edited the question.
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Re: question about remainders  [#permalink]

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New post 19 Feb 2011, 02:26
5
5
144144 wrote:
jeeteshsingh wrote:
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D


can someone plz explain me this other way? thanks.


Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157

If you divide 7^131 by 5, which remainder do you get?
A. 0
B. 1
C. 2
D. 3
E. 4

Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3.

Answer: D.

Hope it's clear.
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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 04:31
2
4
whenever you come across a question like this :-- a number raised to higher power of integer divided by 2,3,4,5,6,7,8,9..
and asking for remainder...then, the question is actually asking for the unit digit of the numerator..

now,
3^4 has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 3.

6 to the power anything, always has a unit digit of 6

7^4 also has a unit digit of 1 => unit digit of 1 will repeat after every fourth power of 7

now, 151 = 148 + 3

7^148 * 7^3 = unit digit of 1 * unit digit of 3 = unit digit of 3..

any number divided by 5 gives remainder = unit digit of the number, when unit digit is less than 5

and (unit digit - 5) , when unit digit is greator than 5
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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 03:18
1
2
Same as asking

If you divide 7^3 by 5, which remainder do you get?
hence OA must be d) 3

because 131=4(K) + 3

now where from I am getting these
follow the pattern

when you place 1,you get 7^1 and results in 7 in unit digit

1---7
2---49--hence 9
3---343---hence 3
4--1


now its repeated


5-7
6-9
7-3
8-1
so on..

131=128+3=4(32)+3

hope i am making sense:)


blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?

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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 07:50
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


If you look at the power of 7, it shows a repeated trend at the Unit's digit. For e.g.:
7^1 = 7 (7 at unit place)
7^2 = 49 (9 at unit place)
7^3 = 343 (3 at unit place)
7^4 = 2401 (1 at unit place)
7^5 = 16807 (7 at unit place)

So if you see, this trend of 7,9,3,1......7,9,3,1.......repeats itself.
7^131 will give 3 as unit's digit and when it is divided by 5, it will give the remainder as 3.

Hope it helps.
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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 08:27
2
when you divide 7 by 5 we get 2 as remainder

so now we can write it as 7^131 \ 5 = 2 ^131 \ 5

= 2^ 120 * 2^ 11 \ 5
= (2^4)^30 * 2048 ( when we multiply 2^11 we will get the value as 2048)
= 16^30 * 2048
= 1^ 30 * 2048 ( when we divide 16 by 5 the remainder is 1 so 1^30 = 1

finally divide 2048 by 5 and u get the remainder = 3

hence 1 * 3 = 3

So answer is D
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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 08:36
5
there is another easy method

cyclicity of 7 = 4

so when we divide 131 by 4 we get 3 as remainder

therefore 7^3 = 343

now divide 343 by 5 and u get the remainder as 3

so answer is D........

Note : cyclicity is used to find the last digit of the number i.e in case of powers it is specially usefull
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Re: question about remainders  [#permalink]

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New post 01 Sep 2009, 19:19
7*7*7*7= xxx1 -> only last digit is important.

you can use this for 32 time and get to 7^128 and last digit will be 1.

1 *7*7*7 = 343 and divided by 5 leaves a remainder of 3.
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Re: question about remainders  [#permalink]

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New post 31 Jan 2010, 06:24
1
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

a) 0
b) 1
c) 2
d) 3
e) 4

I can't get question right, can anyone help me?


7^131 Mod 5
= (7 Mod 5) ^ 131
= 2 ^ 131
= (2 Mod 5)^ 130 * (2 Mod 5)
= (4 Mod 5)^65 * 2
= (-1)^65 * 2
= -2

Therefore remainder = 5-2 = 3 (since -2 obtained is negative)

Therefore D
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Re: question about remainders  [#permalink]

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New post 07 Dec 2010, 10:18
7^1 = 7 => 7/5 has R2
7^2 = 49 => 49/5 has R4
7^3 = 343 => 343/5 has R3
7^4 = 2401 =>2401/5 has R1
7^5 = 16807 => 16807/5 has R2

So it repeats every 4. 131/4 = 128 R3
So 7^128 / 5 has the same remainder as 7^4 /5
7^129 / 5 has the same remainder as 7^1 / 5
7^130 / 5 has the same remainder as 7^2 / 5
7^131 / 5 has the same remainder as 7^3 / 5, which is 3.

The answer is 3.
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If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 09 Mar 2014, 12:08
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Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 01 Apr 2014, 04:37
Actually in my sleep yesterday it occurred to me that something such as the following could be done.
Please advice is this method is not indeed flawed

Whenever we have 5 in the denominator we can multiply both numerator and denominator by 2 so we get 10 in the denominator and therefore we can just find the units digit in the numerator

It would be something like this

7^131/5 = 7^131*2/10

7 has cyclisity of 7,9,3,1 therefore units digit is 3*2=6 but since we multiplied by 2, then 3

Answer is thus 3

* Remember that the remainder of a number when divided by 2 is the units digit. Likewise, the remainder of a number when divided by 100 are the last two digits

Hope it makes sense
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Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 01 Apr 2014, 21:31
jlgdr wrote:
Actually in my sleep yesterday it occurred to me that something such as the following could be done.
Please advice is this method is not indeed flawed

Whenever we have 5 in the denominator we can multiply both numerator and denominator by 2 so we get 10 in the denominator and therefore we can just find the units digit in the numerator

It would be something like this

7^131/5 = 7^131*2/10

7 has cyclisity of 7,9,3,1 therefore units digit is 3*2=6 but since we multiplied by 2, then 3

Answer is thus 3

* Remember that the remainder of a number when divided by 2 is the units digit. Likewise, the remainder of a number when divided by 100 are the last two digits

Hope it makes sense
Cheers
J


Your method is correct but for this question. If you generalize it, it could be flawed. The reason is this: if there is a number with units digit as 6 (e.g. .......6), when you divide it by 2, the last digit could be 3 but it could also be 8. Here we know that we multiplied a power of 7 so the last digit CANNOT be 8 so your method is fine but be careful when you try to generalize it.

Also, you don't need to multiply by 2 to make the denominator 10. Even when the denominator is 5, the last digit is enough to give you the remainder. If the last digit is from 0 to 4, the remainder is the same as the last digit. If the last digit is from 5 to 9, remainder is (last digit - 5).
7^131 ends with 3 so remainder when divided by 5 must be 3.
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Re: question about remainders  [#permalink]

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New post 29 Apr 2014, 10:36
Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157

If you divide 7^131 by 5, which remainder do you get?
A. 0
B. 1
C. 2
D. 3
E. 4

Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3.

Answer: D.

Hope it's clear.[/quote]

bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller
integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3

thanks :)
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Re: question about remainders  [#permalink]

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New post 30 Apr 2014, 06:41
adymehta29 wrote:

bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller
integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3

thanks :)


Yes, 3 divided by 5 gives the remainder o f 3. I don't understand the rest of your post though...
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Re: question about remainders  [#permalink]

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New post 30 Apr 2014, 22:27
1
adymehta29 wrote:
Check this first: 218-if-x-and-y-are-positive-integers-what-is-the-remainder-109636.html#p875157

If you divide 7^131 by 5, which remainder do you get?
A. 0
B. 1
C. 2
D. 3
E. 4

Last digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

As the remainder upon division 131 by 4 (cyclicity) is 3 then the last digit of 7^131 is the same as that of 7^3 so 3 (the third digit from the pattern {7, 9, 3, 1}). Now, any positive integer ending with 3 upon division by 5 yields the remainder of 3.

Answer: D.

Hope it's clear.

bunuel, as per the theory in gmat club math book "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller
integer." i tried solving the above sum and got the correct answer just want to know whether the method is correct or not , as 3 is smaller than 5 so the remainder would be 3

thanks :)


Hey Bunuel,

If you don't mind, I think I see the problem Ady is facing so I will take it up.

Note that the gmat club math book says "When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller
integer."

So when 3 is divided by 5, remainder is 3. When 10 is divided by 39, remainder is 10 and so on...

But in this question, you have \(7^{131}\) divided by 5. You find that \(7^{131}\) ends in 3. This means it is a number which looks something like this:
\(7^{131} = 510320.....75683\) (a huge number that ends in 3. Other than 3, I have used some random digits.)

So the remainder is 3 not because 3 is smaller than 5. The actual number is much much bigger than 5. For example, if you have 276543 divided by 5, will you say that the number is smaller than 5 and hence the remainder is 3? No. The remainder is 3 because the number ends in 3 and when you divide such a number by 5, you know that the number which is 3 steps before it i.e. 276540 in this case, is a multiple of 5 (every number ending in 0 or 5 is a multiple of 5). That is the reason that you will be left with 3 when you divide this number by 5.

Similarly, 4367 divided by 5 leaves remainder 2 because 4365 is divisible by 5 and so on...

The first part of this post discusses remainders in case of division by 5: http://www.veritasprep.com/blog/2014/03 ... emainders/
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If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 27 Mar 2018, 23:51
hi guys, the cycle is 7,9,3,1 and unite digit is 3 here so when we divide 3/5 the remainder is 3, how come it comes 3? please explain
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Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 28 Mar 2018, 01:35
1
imtisalsiddiqui wrote:
hi guys, the cycle is 7,9,3,1 and unite digit is 3 here so when we divide 3/5 the remainder is 3, how come it comes 3? please explain


Your question is not clear. Any positive integer ending with 3 upon division by 5 yields the remainder of 3:

3 divided by 5 gives the remainder of 3;
13 divided by 5 gives the remainder of 3;
23 divided by 5 gives the remainder of 3;
...
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Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 28 Mar 2018, 02:09
Thanks for clearing up my confusion, understood!
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Re: If you divide 7^131 by 5, which remainder do you get?  [#permalink]

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New post 29 Mar 2018, 22:55
blackcrow wrote:
If you divide 7^131 by 5, which remainder do you get?

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


7 has cyclicity of 4 = 7, 9, 3, 1

Dividing 131 by 4, we get remainder 3

In cyclicity, third term we have is "3"

Hence (D)
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Re: If you divide 7^131 by 5, which remainder do you get? &nbs [#permalink] 29 Mar 2018, 22:55

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