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by Night reader » Fri Feb 11, 2011 4:44 am
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
IOM E in a series of {4,11,18,25,32,39,46,53,60,67,74,81} :)

3(n-1)/7=i (i is integer) +2/7 as 5+r should give us divisibility by 7; 3(n-1)=7i+2 --> 3n-3-2=7i <> 3n-5=7i
start from n=4, continue until (n+7)<=85
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by GMATGuruNY » Fri Feb 11, 2011 6:59 am
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
Let's determine the pattern.
n=1: 5 + (1-1)*0 = 5.
n=2: 5 + (2-1)*3 = 8.
n=3: 5 + (3-1)*3 = 11.
n=4: 5 + (4-1)*3 = 14. 14 is a multiple of 7.

The first multiple of 7 occurs when n=4.
We can see that each number in the sequence is 3 more than the previous number in the sequence.
This means that every 7th term increases by 7*3 = 21.
Since n=4 is a multiple of 7, and every 7th term after n=4 will increase by 21, every 7th term after n=4 will be a multiple of 7:

n=11: 5 + (11-1)*3 = 35.
n=18: 5 + (18-1)*3 = 56.
And so on.

Since 85-4 = 81, and 11*7 = 77, there are 11 multiples of 7 between n=4 and n=85.
Thus, including n=4, the total number of multiples of 7 = 1+11 = 12.

The correct answer is E.
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by gtestprep » Fri Feb 11, 2011 8:51 am
Thanks GMATGuruNY! Your explanation helped.
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by viv_gmat » Fri Nov 18, 2011 10:29 am
GMATGuruNY wrote:
gtestprep wrote:Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85 (inclusive). How many of these numbers are divisible by 7?

A. 5
B. 7
C. 8
D. 11
E. 12
Let's determine the pattern.
n=1: 5 + (1-1)*0 = 5.
n=2: 5 + (2-1)*3 = 8.
n=3: 5 + (3-1)*3 = 11.
n=4: 5 + (4-1)*3 = 14. 14 is a multiple of 7.

The first multiple of 7 occurs when n=4.
We can see that each number in the sequence is 3 more than the previous number in the sequence.
This means that every 7th term increases by 7*3 = 21.
Since n=4 is a multiple of 7, and every 7th term after n=4 will increase by 21, every 7th term after n=4 will be a multiple of 7:

n=11: 5 + (11-1)*3 = 35.
n=18: 5 + (18-1)*3 = 56.
And so on.

Since 85-4 = 81, and 11*7 = 77, there are 11 multiples of 7 between n=4 and n=85.
Thus, including n=4, the total number of multiples of 7 = 1+11 = 12.

The correct answer is E.

Hi,
Is there a short cut to solve this problem?

Your prompt reply will be highly appreciated.

Thanks
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by vishal.pathak » Sat Nov 19, 2011 5:38 am
viv_gmat wrote: Hi,
Is there a short cut to solve this problem?

Your prompt reply will be highly appreciated.

Thanks
We have to determine the numbers divisible by 7
eqn is 5+(n-1)*3
So (n-1)*3 should give us a remainder of 2 when divided by 7.
We know that 9 (3*3) gives a remainder of 2 when divided by 7
Every 7th multiple of 3 will give a remainder of 2 (like 10 (3+7) and 17 (10+7) etc)
Number of such multiples in 84 (85-1) = 84/7 = 12
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by KItuz » Tue Mar 20, 2012 7:09 am
Question can be reduced to how many integers n belong to [1,...,85] such that (3n+2)/7 is an integer.
multiples of 7 are 7,14,21,28,35,.....
subtracting 2 from each we have 5,12,19,26,33,....
expressing each of them as 3*x we have x,3*4,x,x,3*11,...
Blind guess - checking if there is a pattern here we get 11-4=7
so the question can be reduced to count m such that 4+7m<=85
substitute 12 and 4+7m >85 so we have 0,1,2,3,4,...11 here = 12 values.
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by ronnie1985 » Wed Mar 21, 2012 9:48 am
The expression is 3n+2. When divided by 7:-

n= even = 2k, remainder = 2-k
n= odd = 2k-1, remainder = 1-k
For n = even, the numbers shall be divisible for all 2-k = 0,-7,-14,-21,-28 = > k = 2, 9, 16, 23, 30, 37 = > n = 4, 18, 32, 46, 60, 74.
For n = odd, the numbers shall be divisible for all 1-k = 0, -7, -14,-21,-28, -35,...etc.
=> n = 8, 15, 22, 29, 36, 43.
Total = 12 such n. (E) is answer.
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by anjalimanas » Fri Mar 15, 2013 1:29 am
Alternate way

Multiples of 7 from 5 to 257 = 36

as the series is advancing by 3 everytime, Number of multiples of 7 in series = 36 / 3 = 12
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