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

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]
01 Apr 2014, 21:31

Expert's post

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. _________________

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

Re: question about remainders [#permalink]
30 Apr 2014, 06:41

Expert's post

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... _________________

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...

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]
21 May 2014, 06:19

If we raise the power of 7 by 2 , last digit will be 9 (7X7 = 49) If we raise the power of 7 by 3, last digit will be 3 (7X7X7 = 343) If we raise the power of 7 by 4 , last digit will be 1 (A9 X A9 = XYZ1; Here A is 49) If we raise the power of 7 by multiple of 4 , last digit will be always 1 .

Now 131 = 4X32 + 3

Raising the power of 7 by 131 is equal to raising the power by 4X32 and then multiplying by 7^3

last digit of first term will be 1 and second term will be 3

Hence last digit of 7^131 will be 1X3 = 3

Since last digit is less than 5, if we divide the number by 5 we will get 3 as remainder ( If last digit is 5 or greater than 5 we have to subtract 5 from last digit to get the remainder)

Re: If you divide 7^131 by 5, which remainder do you get? [#permalink]
20 Oct 2014, 01:25

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

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

We can solve this problem using the two ways:

1. This method is limited to this question or any question which has 5 or 10 or their power in the denominator.

If I can know the last 2 digits of this, I can know what will be the remainder of 7^131 when divided by 5, by dividing the last 2 digits by 5.

To calculate the last two digits, we need to see the pattern of the digits - 7 49 43 01 07 49

The cyclicity is 4. Now dividing 131 by cliclicity to know the 2 digit number. The remainder is 3 thus the last 3 digits will be 03.

Now a number with last 2 digits as 03, will give remainder 3 when divided by 5.

Thus 3.

The second method is generic and will work with all the possible scenarios.

When 7 is divided by 5, the remainder is 2. 2 has a cyclicity of 4 with 131, we'll be left with 3. 2^3 = 8 when we divide this by 5, we get the remainder 3.

Thus Ans = 3 (D) _________________

Give KUDOS if the post helps you...

gmatclubot

Re: If you divide 7^131 by 5, which remainder do you get?
[#permalink]
20 Oct 2014, 01:25

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