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# When positive integer n is divided by 3, the remainder is 2. When n is

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Intern
Joined: 25 Sep 2016
Posts: 16
Re: When positive integer n is divided by 3, the remainder is 2. When n is  [#permalink]

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03 Mar 2017, 13:43
1
n=3p+2 (2,5,8,11,14,17,20,23,26....)
n=7q+5 (5,12,19,26,41,68,89....)

hence maximum no. N can take is 5.
so answer is 'E'.
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When positive integer n is divided by 3, the remainder is 2. When n is  [#permalink]

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17 Aug 2017, 10:10
Ans is E

n= 3i + 2
series will be 2,5,8,11..............26.......47.....68.....89......

also n= 7i +5
series will be 5,12,19,26,.........47.......68......89.....

taking common from both series

5 numbers are = 5,26,47,68,89
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Re: When positive integer n is divided by 3, the remainder is 2. When n is  [#permalink]

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21 Aug 2017, 16:41
Bunuel wrote:
When positive integer n is divided by 3, the remainder is 2. When n is divided by 7, the remainder is 5. How many values less than 100 can n take?

(A) 0
(B) 2
(C) 3
(D) 4
(E) 5

Kudos for a correct solution.

Since when n is divided by 3 the remainder is 2, n can be 2, 5, 8, 11, 14, 17, 20, 23, 26, etc.

Since when n is divided by 7 the remainder is 5, n can be 5, 12, 19, 26, etc.

We see that 5 is the first match for n. The next possible value of n is 5 + 7 x 3 = 26, then 26 + 21 = 47, then 47 + 21 = 68, and finally 68 + 21 = 89.

So, n can be 5, 26, 47, 68, or 89.

Alternate Solution:

Since the remainder from the division of n by 3 is 2, n = 3p + 2 for some integer p.

Since the remainder from the division of n by 7 is 5, n = 7q + 5 for some integer q.

Notice that n - 5 = 3p - 3 = 7q is divisible by 3 and 7; therefore, n - 5 is divisible by 21.

Thus, n - 5 could be 0, 21, 42, 63, or 84.

Thus, n could be 5, 26, 47, 68, or 89.

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Re: When positive integer n is divided by 3, the remainder is 2. When n is  [#permalink]

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20 Sep 2019, 20:24
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Re: When positive integer n is divided by 3, the remainder is 2. When n is   [#permalink] 20 Sep 2019, 20:24

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# When positive integer n is divided by 3, the remainder is 2. When n is

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