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If n is the remainder when 250 is divided by 3 and m is the remainder

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If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 27 Dec 2016, 08:01
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Re: If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 27 Dec 2016, 09:53
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Bunuel wrote:
If n is the remainder when 2^50 is divided by 3 and m is the remainder when 2^15 is divided by 5, what is m + n?

A. 6
B. 5
C. 4
D. 3
E. 2


2 = -1 (mod 3)

\(\frac{2^{50}}{3} = \frac{(-1)^{50}}{3} = \frac{1}{3}\) remainder 1

\(2^4\) = 1 (mod 5)

\(\frac{(2^4)^3*2^3}{5} = \frac{1*8}{5}\) remainder 3

1 + 3 = 4

Answer C
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If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 27 Dec 2016, 10:41
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There is repetitive pattern for remainders.
(2^1)/3 - Remainder is 2
(2^2)/3 - Remainder is 1
(2^3)/3 - Remainder is 2, and so on.
So for even powers, remainder is 1. Therefore remainder for (2^50)/3 is 1.

Similarly, we can observe a pattern for remainders when divided by 5.
(2^1)/5 - Remainder is 2
(2^2)/5 - Remainder is 4
(2^3)/5 - Remainder is 3
(2^4)/5 - Remainder is 1
(2^5)/5 - Remainder is 2
(2^6)/5 - Remainder is 4
(2^7)/5 - Remainder is 3
We can see that there is a pattern of 2,4,3,1. So remainder for (2^15)/5 is 3.

Therefore m+n = 1+3 = 4.

Option C is the right answer.


This is my first answer on a forum. If it helped you out, please leave a kudos. :)
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Re: If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 27 Dec 2016, 10:53
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Bunuel wrote:
If n is the remainder when 2^50 is divided by 3 and m is the remainder when 2^15 is divided by 5, what is m + n?

A. 6
B. 5
C. 4
D. 3
E. 2


\(\frac{2^4}{3}\) = Remainder 1
So, \(2^{50} = 2^{4*12}*2^2\)

Now, \(\frac{2^2}{3}\) = Remainder 1

So, m = 1

\(\frac{2^4}{5}\) = Remainder 1
So, \(2^{15} = 2^{4*3}*2^3\)

\(\frac{2^3}{5}\) = Remainder 3

So, n = 3

Hence, \(m + n = 1 + 3 = 4\)

Hence, Correct answer will be (C) 4
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Re: If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 15 Jan 2017, 03:03
Hi,

I don´t quite understand the following


243243 = Remainder 1
So, 250=24∗12∗22250=24∗12∗22

Now, 223223 = Remainder 1

So, m = 1

245245 = Remainder 1
So, 215=24∗3∗23215=24∗3∗23

235235 = Remainder 3

So, n = 3

Hence, m+n=1+3=4m+n=1+3=4


Can anyone expand the explanation?

Thanks
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Re: If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 29 Mar 2017, 05:12
KatariaP wrote:
There is repetitive pattern for remainders.
(2^1)/3 - Remainder is 2
(2^2)/3 - Remainder is 1
(2^3)/3 - Remainder is 2, and so on.
So for even powers, remainder is 1. Therefore remainder for (2^50)/3 is 1.

Similarly, we can observe a pattern for remainders when divided by 5.
(2^1)/5 - Remainder is 2
(2^2)/5 - Remainder is 4
(2^3)/5 - Remainder is 3
(2^4)/5 - Remainder is 1
(2^5)/5 - Remainder is 2
(2^6)/5 - Remainder is 4
(2^7)/5 - Remainder is 3
We can see that there is a pattern of 2,4,3,1. So remainder for (2^15)/5 is 3.

Therefore m+n = 1+3 = 4.

Option C is the right answer.


This is my first answer on a forum. If it helped you out, please leave a kudos. :)


Great solution! Thanks!
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Re: If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 31 Mar 2017, 11:33
Bunuel wrote:
If n is the remainder when 2^50 is divided by 3 and m is the remainder when 2^15 is divided by 5, what is m + n?

A. 6
B. 5
C. 4
D. 3
E. 2


Let’s determine the remainder pattern when 2 raised to an exponent is divided by 3.

(2^1)/3 = 0 remainder 2

(2^2)/3 = 4/3 = 1 remainder 1

(2^3)/3 = 8/3 = 2 remainder 2

(2^4)/3 = 16/3 = 5 remainder 1

When 2 is raised to an even exponent and is divided by 3, a remainder of 1 results. Thus, 2^50 divided by 3 has a remainder of 1.

Next we can determine the remainder when 2^15 is divided by 5. To do so, we recall that the units digit of any number divided by 5 will produce the same remainder as when the actual number is divided by 5. Thus, let’s determine the units digit of 2^15. The pattern of units digits of the base of 2 when raised to an exponent is:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 6

2^5 = 2

We see the pattern is 2-4-8-6, and furthermore that 2^4k, in which k is a positive integer, will always have a units digit of 6.

Thus, the units digit of 2^16 = 6 and so the units digit of 2^15 = 8. Dividing 8 by 5 yields a remainder of 3; thus, dividing 2^15 by 5 also yields a remainder of 3.

Therefore, m + n = 3 + 1 = 4.

Answer: C
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If n is the remainder when 250 is divided by 3 and m is the remainder [#permalink]

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New post 31 Mar 2017, 13:13
Bunuel wrote:
If n is the remainder when 2^50 is divided by 3 and m is the remainder when 2^15 is divided by 5, what is m + n?

A. 6
B. 5
C. 4
D. 3
E. 2


Solution


Solution



    • When \(2^{50}\) is divided by \(3\) the remainder is n. Let us find this “\(n\)”.

      o \(\frac{2^{50}}{3} = \frac{(2^2)^{25}}{30} = \frac{4^{25}}{3}\)

      o We know when 4 is divided by 3, the remainder is 1.

      o Thus \(n = [\frac{4^{25}}{3}]_R = (1^{25}) = 1\)

    • When \(2^{15}\) is divided by \(5\), the remainder is \(m\).

      o \(\frac{2^{15}}{5} = \frac{[(2^4)^3 * 2^3]}{5}\)

      o We know that when 16 is divided by 5 the remainder 1 and 8 divided by 5, the remainder is 3

      o Thus, \(m = [\frac{16^3}{5}]_R * [\frac{8}{5}]_R = 1 * 3 = 3\)

    • Therefore, \(m+n = 1 + 3 = 4\)

    • Correct answer is Option C.


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Saquib
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If n is the remainder when 250 is divided by 3 and m is the remainder   [#permalink] 31 Mar 2017, 13:13
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