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Temurkhon
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Consider a sequence of numbers given by the expression 5 + ( 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
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vitaliyGMAT
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Consider a sequence of numbers given by the expression 5 + ( 04 Nov 2016, 07:14
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Marcab
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
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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.
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Consider a sequence of numbers given by the expression 5 + ( 21 Aug 2017, 00:00
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Marcab
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
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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
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Consider a sequence of numbers given by the expression 5 + ( 06 Sep 2017, 06:58
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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
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Consider a sequence of numbers given by the expression 5 + ( 12 Sep 2017, 19:21
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Bunuel
Marcab
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.

Answer: E.

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?
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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
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Consider a sequence of numbers given by the expression 5 + ( 08 May 2021, 07:56
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Sequence is 5 + (n - 1) * 3 = 3n +2
So, Nos will be 5, 8, 11, 14, 17...... 254, 257 But for divisible by 7. n should be as
\(t_1\) = 4, \(t_2\) = 11, \(t_3\)= 11,....... \(t_85\)= 81

Nos. of terms =\(\frac{ last term- first term }{7} +1\) =(81 - 4)/7 +1 =12.

So, It is E. :)
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Consider a sequence of numbers given by the expression 5 + ( 04 Aug 2026, 03:12
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First I find 2 values divisible by 7
5+(n-1)3 = 3n+2
2,5,8,11,14,17,20,23,26,29,32,35..... => 14 and 35 are two numbers divisible by 7 and have a difference of 21 since it is AP we will get next value divisible by 35+21 = 56
Therefore using AP formula to find number of terms
3*85+2 = 14 + (n - 1)21
257 - 14 /21 = n - 1
~ 11 = n -1
n= 12
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