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

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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

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

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]

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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]

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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.

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

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26 Jan 2016, 20:49

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

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26 Jan 2016, 20:56

VeritasPrepKarishma Bunuel I started doing this question a couple of wks after going through remainders and binomial theorem. My first instinct was to use the binomial theorem, and I failed to remember it/do it correctly.

Can you help me understand when I would use binomial theorem, vs. the simple solution that is used here of just finding the pattern?

My suspicion is that binomial theorem will be used if we have a denominator >10. Please let me know if my understanding is correct. Thank you!!

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