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Temurkhon
Joined: 23 Jan 2013
Last visit: 06 Apr 2019
Posts: 408
Own Kudos:
Given Kudos: 43
Schools: Cambridge'16
07 Jun 2015, 05:18
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we can ignore n*3+5 because it is the same as 3q+r, i.e every n from 1 to 85 should be divisible by 7, so it is 7 to 84.
(84-7/7)+1=12
E
(84-7/7)+1=12
E
04 Nov 2016, 07:14
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It’s relatively fast and easy to use modular arithmetic in solving this question
The given arithmetic sequence should be divisible by 7 and we need to find the values of \(n\) which will satisfy this.
So we have:
5 + (n-1)*3 ≡ 0 (mod 7)
(n-1)*3 ≡ -5 (mod 7)
(n-1)*3 ≡ 9 (mod 7)
n-1 ≡ 3 (mod 7)
n ≡ 4 (mod 7)
Our \(n\) leaves remainder 4 when divided by 7 and has following form:
\(n = 7*x+4\)
This is arithmetic progression which starts with 4 and ends with 81 with common difference 7.
Total number of elements in it is:
\(\frac{(81-4)}{7}+1=12\)
Answer E.
21 Aug 2017, 00:00
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We start putting values of n from 1 to 85 and find that
n exp.
1 5
2 8
3 11
4 14
....................
....................
11 35
.....................
......................
18 56
So, The numbers are divisible by 7 for n = 4,11,18, 25, .............., 81
From formula of AP , 81 = 4 +(n-1)7
n = 77/7 + 1 = 12
Answer E
