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When positive integer k is divided by 5, the remainder is 2. When k is

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When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 25 Oct 2010, 11:08
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When positive integer k is divided by 5, the remainder is 2. When k is divided by 6, the remainder is 5. If k is less than 40, what is the remainder when k is divided by 7?

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

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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 25 Oct 2010, 11:24
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shrive555 wrote:
When positive integer k is divided by 5, the remainder is 2. When k is divided by 6, the remainder is 5. If k is less than 40, what is the remainder when k is divided by 7?


2
3
4
5
6

Any algebraic approach to this question


Cant think of a straight approach but here is how I solved it:
K is divided by 5 and remainder is 2. This means k = 5n + 2 (n is an integer)
so the possible values of K = {2, 7, 12, 17, 22, 27, 32, 37} (less than 40)
Secondly, if K is divided by 6, the remainder is 5 => k= 6m + 5
so the possible value set for k = {5, 11, 17, 23, 29,35} (less than 40)

17 is the only common number in both the sets. Hence k = 17.
so 3 is the answer.
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 26 Oct 2010, 07:54
17 is the number and the remainder is 3. Answer B
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 26 Oct 2010, 15:52
I dont think there is any easy algebraic way to solve this, the best approach is to enumerate out the possibilities and eliminate to get the answer as highlighted in the solution above. That is why you are given the constraint of the answer being less than 40, to make this search & elimination easier.
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 26 Oct 2010, 21:48
shrouded1 wrote:
I dont think there is any easy algebraic way to solve this, the best approach is to enumerate out the possibilities and eliminate to get the answer as highlighted in the solution above. That is why you are given the constraint of the answer being less than 40, to make this search & elimination easier.


Thanks. i guess that's a good clue. the constraint of 40.
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 15 Mar 2018, 05:26
shrive555 wrote:
When positive integer k is divided by 5, the remainder is 2. When k is divided by 6, the remainder is 5. If k is less than 40, what is the remainder when k is divided by 7?

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


If we list out the numbers < 40, than 17 is the only common number for both 5 and 6 that gives the remainder 2 & 5 respectively.

Therefore, 17/ 7 gives remainder 3

(B)
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 19 Mar 2018, 06:37
shrive555 wrote:
When positive integer k is divided by 5, the remainder is 2. When k is divided by 6, the remainder is 5. If k is less than 40, what is the remainder when k is divided by 7?

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


We are given that k < 40. Since when positive integer k is divided by 5, the remainder is 2:

k = 5Q + 2

So k can be 2, 7, 12, 17, 22, 27, 32, or 37.

Since, when k is divided by 6, the remainder is 5:

k = 6Q + 5

So k can be 5, 11, 17, 23, 29, or 35.

Thus, we see that k must be 17, and 17/7 = 2 remainder 3.

Answer: B
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 08 Apr 2018, 07:18
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Cant think of a straight approach but here is how I solved it:
K is divided by 5 and remainder is 2. This means k = 5n + 2 (n is an integer)
so the possible values of K = {2, 7, 12, 17, 22, 27, 32, 37} (less than 40)
Secondly, if K is divided by 6, the remainder is 5 => k= 6m + 5
so the possible value set for k = {5, 11, 17, 23, 29,35} (less than 40)

17 is the only common number in both the sets. Hence k = 17.
so 3 is the answer.
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When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post Updated on: 18 Sep 2018, 14:25
shrive555 wrote:
When positive integer k is divided by 5, the remainder is 2. When k is divided by 6, the remainder is 5. If k is less than 40, what is the remainder when k is divided by 7?

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


k=5q+2
k=6p+5
5q+2=6p+5➡
5q/3=2p+1
because 2p+1 is odd, q must be odd multiple of 3
least value of q that will make p an integer is 3
k=5*3+2=17
17/7 gives a remainder of 3
B

Originally posted by gracie on 08 Apr 2018, 11:46.
Last edited by gracie on 18 Sep 2018, 14:25, edited 1 time in total.
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Re: When positive integer k is divided by 5, the remainder is 2. When k is  [#permalink]

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New post 18 Sep 2018, 01:42
Dividend = Divisor * Quotient + Remainder.
K=5Q+2 ----------(1)
K=6Q+5 ----------(2)
K=5Q+Q+5
K=(5Q+5)+Q
K=5(Q+1)+Q -----------(3)
Comparing (1) and (3) we get -->
Q = 2.
Therefore, K=(6*2)+5 = 17.
K/7 = 17/7 = 3(Remainder).
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Re: When positive integer k is divided by 5, the remainder is 2. When k is &nbs [#permalink] 18 Sep 2018, 01:42
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