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08 Dec 2017, 02:57
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I was just fooling around with remainders and had a query.
Suppose I want to find the remainder when \(7^{30}\) is divided by 100. I know I can solve this using cyclicity, but why don't I get the right answer using the following method?
I can express \((7^{30})/100\) as \([(7^{15})/10]^{2}\). Now why can't I find the remainder of \((7^{15})/10\) and simply square it to obtain the desired answer?
(Answer is 49)
Suppose I want to find the remainder when \(7^{30}\) is divided by 100. I know I can solve this using cyclicity, but why don't I get the right answer using the following method?
I can express \((7^{30})/100\) as \([(7^{15})/10]^{2}\). Now why can't I find the remainder of \((7^{15})/10\) and simply square it to obtain the desired answer?
(Answer is 49)
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08 Dec 2017, 07:13
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dejavu619
hi...
by this method you are NOT getting the remainder because you are changing the DIVISOR from 100 to 10
if you were to find the remainder when 10 is divisor it is OK..
\(\frac{7^{30}}{10}=\frac{(7^{15})^2}{10}\)
Now you find remainder of 7^15 and the square it..
A very small example is..
remainder of 49 when divided by 36.. ans 49-36=13
but if you look at it as SQUARE, 7^2 divided by 6^2..
\(\frac{7^2}{6^2}=(\frac{7}{6})^2\) so 1^2=1.. NO
30 Dec 2017, 01:03
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dejavu619
We know that any number can be written in terms of its Divisor, Quotient and Remainder.
Thus, we can write \(N = QD + R\) ......(i)
According to your logic, if the remainder when N divided by D is R, then the remainder when \(N^2\) is divided by \(D^2\) should be \(R^2\).
If you square equation (i), you will be able to see it clearly that your assumption is not correct.
\(N^2 = (QD + R)^2\)
\(N^2 = Q^2* D^2 + R^2 + 2 * Q * R * D\)
Keep in mind that you are dividing \(N^2\) by \(D^2\) now..
\(Q^2 * D^2\) is diivisble by \(D^2\).
But what about \(2*Q * R * D\), do we know if this is perfectly divisble by \(D^2\)?
No, we don't!
Hence, it is wrong to make such assumption that if \(N/D\) give \(R\) as remainder then \(N^2/D^2\) will give \(R^2\) as the remainder or vice versa.
Regards,
Saquib
e-GMAT
Quant Expert
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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Thank you for understanding, and happy exploring!
