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

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

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:)

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

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.

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

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... emainders/

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;
...

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

Yes, it is.
Note that when dividing by 5, the units digit alone determines the remainder. Since cyclicity of units digit of both 7 and 2 is 4, we know without calculating that cyclicity of remainders will be 4 too. Hence we don't even need to use binomial in the first step.

Check these posts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... questions/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... ns-part-2/

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.

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.

For more on this kind of questions check Units digits, exponents, remainders problems collection.

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.

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

Yes, 3 divided by 5 gives the remainder o f 3. I don't understand the rest of your post though...

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