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Director  G
Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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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
Senior Manager  B
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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1
Marcab wrote:
Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12

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:

$$\frac{(81-4)}{7}+1=12$$

Director  D
Joined: 13 Mar 2017
Posts: 728
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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Marcab wrote:
Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12

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
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Manager  G
Joined: 27 Jan 2016
Posts: 123
Schools: ISB '18
GMAT 1: 700 Q50 V34 Re: Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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5+(n-1)3 = 3n+2 -> the number when divided by 3 leaves a reminder 2.

The value shoould be divisible by 7 -> the number when divided by 7 leaves a reminder.

Now, the question can be rephrased as find the number of values that when divided by 3 and 7 leave the reminders 2 and 0 respectively.

So 14+(n-1)3 <257
-> n=12
Manager  B
Joined: 19 Aug 2016
Posts: 75
Re: Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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Amateur wrote:
Bunuel wrote:
Marcab wrote:
Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12

We want $$5 + (n - 1) * 3=3n+2$$ to be divisible by 7, which happens when n is 4, 11, 18, ..., 81. Total of 12 numbers.

P.S. Please provide OA's for the questions.

hey Bunuel, is there a way to cut down the time for counting those 12 numbers?

Yes, there is a pattern

When you simplify u will get 3n+2

3n+2=5
3n+2=8
3n+2=11
3n+2=14...for values 1 2 3 4 respectively

from numbers 1 to 85..84 is divisible by 7

84/12=7 so 12 numbers
Non-Human User Joined: 09 Sep 2013
Posts: 13136
Re: Consider a sequence of numbers given by the expression 5 + (  [#permalink]

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_________________ Re: Consider a sequence of numbers given by the expression 5 + (   [#permalink] 14 Feb 2019, 22:38

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