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aiming4mba
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Another remainder question 21 Mar 2011, 12:44
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What is the remainder when 7^84/342

A. 1
B. 2
C. 3
D. 4
E. 5



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Another remainder question 22 Mar 2011, 18:32
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aiming4mba
What is the remainder when 7^84/342

A. 1
B. 2
C. 3
D. 4
E. 5
Show more

First up, GMAT questions don't involve painful calculations. So there has to be some obvious link between 7 and 342. It helps one to know the squares of first 20 numbers and cubes of first 10 numbers.
What immediately comes to mind is that 7^3 = 343

So \(\frac{7^{84}}{342} = \frac{(7^3)^{28}}{342} = \frac{343^{28}}{342} = \frac{(342 + 1)^{28}}{342}\)

Now, when you open \((a + b)^n\), every term will be divisible by a except the last term i.e. \(b^n\) (using binomial theorem)
So every term of \((342 + 1)^{28}\) will be divisible by 342 except the last term 1^{28} = 1. Hence remainder is 1.

If you are not comfortable with binomial theorem, don't worry about it. Just remember that when you write down all the terms of \((a + b)^n\), each term is divisible by a except the last term b^n. To see an example, let's take a small value of n
\((a+b)^2 = a^2 + 2ab + b^2\)
\((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
All terms are divisible by a except the last term.
There are other ways to get to the answer but they are a little cumbersome...

Is this question from some GMAT specific source?
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Another remainder question 22 Mar 2011, 23:31
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Thanks Karishma, its clear now. yes, this question is from a prep material by a coaching centre for GMAT
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Another remainder question 23 Mar 2011, 17:05
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aiming4mba
What is the remainder when 7^84/342

A. 1
B. 2
C. 3
D. 4
E. 5

First up, GMAT questions don't involve painful calculations. So there has to be some obvious link between 7 and 342. It helps one to know the squares of first 20 numbers and cubes of first 10 numbers.
What immediately comes to mind is that 7^3 = 343

So \(\frac{7^{84}}{342} = \frac{(7^3)^{28}}{342} = \frac{343^{28}}{342} = \frac{(342 + 1)^{28}}{342}\)

Now, when you open \((a + b)^n\), every term will be divisible by a except the last term i.e. \(b^n\) (using binomial theorem)
So every term of \((342 + 1)^{28}\) will be divisible by 342 except the last term 1^{28} = 1. Hence remainder is 1.

If you are not comfortable with binomial theorem, don't worry about it. Just remember that when you write down all the terms of \((a + b)^n\), each term is divisible by a except the last term b^n. To see an example, let's take a small value of n
\((a+b)^2 = a^2 + 2ab + b^2\)
\((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
All terms are divisible by a except the last term.
There are other ways to get to the answer but they are a little cumbersome...

Is this question from some GMAT specific source?
Show more
like the way you do these hard remainder problems...
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Another remainder question 25 Mar 2011, 21:56
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VeritasPrepKarishma
aiming4mba
What is the remainder when 7^84/342

A. 1
B. 2
C. 3
D. 4
E. 5

First up, GMAT questions don't involve painful calculations. So there has to be some obvious link between 7 and 342. It helps one to know the squares of first 20 numbers and cubes of first 10 numbers.
What immediately comes to mind is that 7^3 = 343

So \(\frac{7^{84}}{342} = \frac{(7^3)^{28}}{342} = \frac{343^{28}}{342} = \frac{(342 + 1)^{28}}{342}\)

Now, when you open \((a + b)^n\), every term will be divisible by a except the last term i.e. \(b^n\) (using binomial theorem)
So every term of \((342 + 1)^{28}\) will be divisible by 342 except the last term 1^{28} = 1. Hence remainder is 1.
Show more

Really good explanation! I followed your explanation on 2^92/7. There were possible alternatives then. But this one, your logic will work way more faster than what was also suggested there.
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Another remainder question 26 Mar 2011, 19:52
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Banterninja


Really good explanation! I followed your explanation on 2^92/7. There were possible alternatives then. But this one, your logic will work way more faster than what was also suggested there.
Show more

Yes, once you get used to these ideas, you can solve these questions orally.
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Another remainder question 27 Mar 2011, 01:39
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A is alright. But another answer can be 7.

7^84 = (7^4) ^21. i.e 4*21 = 84
7^4 = 2401 will leave remainder 7 when divided by 342.
2401= 7 * 342 + 7

Unit digit of 7^21 is same as the unit digit of 7^1.

aiming4mba
What is the remainder when 7^84/342

A. 1
B. 2
C. 3
D. 4
E. 5
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Another remainder question 27 Mar 2011, 02:53
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Thanks fluke !!!!!. Yeah I realized this when I in the gym :-D Kudos......
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Another remainder question 27 Mar 2011, 03:06
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Brilliant explanation Karishma. +1 to you.
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Another remainder question 04 Apr 2011, 05:02
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Very helpful explanation. I would definitely keep this in mind.

Thank you Karishma...



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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