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If you divide 7^131 by 5, which remainder do you get?
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Updated on: 19 Oct 2012, 03:39
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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
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Originally posted by blackcrow on 01 Sep 2009, 04:06.
Last edited by Bunuel on 19 Oct 2012, 03:39, edited 1 time in total.
Renamed the topic and edited the question.




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Re: question about remainders
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19 Feb 2011, 03:26
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 = 52 = 3 (since 2 obtained is negative) Therefore D can someone plz explain me this other way? thanks. Check this first: 218ifxandyarepositiveintegerswhatistheremainder109636.html#p875157If 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
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01 Sep 2009, 05:31
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
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01 Sep 2009, 04:18
Same as asking If you divide 7^3 by 5, which remainder do you get?hence OA must be d) 3because 131=4(K) + 3now where from I am getting these follow the patternwhen you place 1,you get 7^1 and results in 7 in unit digit 17 249hence 9 3343hence 3 41 now its repeated 57 69 73 81 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
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01 Sep 2009, 08:50
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
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01 Sep 2009, 09:27
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
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01 Sep 2009, 09:36
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
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01 Sep 2009, 20:19
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
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31 Jan 2010, 07:24
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 = 52 = 3 (since 2 obtained is negative) Therefore D
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Re: question about remainders
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07 Dec 2010, 11:18
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?
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09 Mar 2014, 13:08



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Re: If you divide 7^131 by 5, which remainder do you get?
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01 Apr 2014, 05:37
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 Cheers J



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Re: If you divide 7^131 by 5, which remainder do you get?
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01 Apr 2014, 22:31
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
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29 Apr 2014, 11:36
Check this first: 218ifxandyarepositiveintegerswhatistheremainder109636.html#p875157If 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
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30 Apr 2014, 07:41



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Re: question about remainders
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30 Apr 2014, 23:27
adymehta29 wrote: Check this first: 218ifxandyarepositiveintegerswhatistheremainder109636.html#p875157If 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?
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28 Mar 2018, 00:51
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?
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28 Mar 2018, 02:35



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Re: If you divide 7^131 by 5, which remainder do you get?
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28 Mar 2018, 03:09
Thanks for clearing up my confusion, understood!



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Re: If you divide 7^131 by 5, which remainder do you get?
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29 Mar 2018, 23:55
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
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