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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Okay tricky question. My approximation that i made within the 2 mins was:

since n is multiplied by 3 and the result has to be devisable by 7 we can only count multiples of 21. the observed set starts with 5 and ends with 257. since 5 and 257 are not devisable by 7 all numbers are ((257-5)/21)-1=11

but i don't know whether i am correct or not
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Hodor?

Kudo!

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

Put these values of n in the relation.
The numbers which you will be getting on doing so, will be divisible by 7.
Do notice one key point here; these values of n are at a difference of 7.

Moreover I will let you do one experiment in this question. Change the entire relation i.e. let the initial number be anything and let the other part be something like (n-3)*5. Let the range of n be from 1 to 85. IMO the number of multiples in this relation will also come out to be 12.

Do let me know if that helps you.
Good luck
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Let the starting term be 3. Let the range of n be :3<=n<=85.
The relation is 3+(n-3)*5.
Therefore on substituting these values, you will be getting the numbers as 3 8 13 18 23 28 33 38 43 48 53 58 63 68 and bla bla till n=85.

Notice that multiples of 7 appear at regular intervals of 7 i.e. 1st multiple at 6th position,m 2 nd mutiple at 13th position etc.
So in general words, the question is asking the numbr of multiples of till 85, no matter what the relation is.

I implemented this result in another relation as well to get the confirmation. 5 +(n-2)*5--->5 10 15 20 25 30 35 40 45 50 55 60 65 70

Hope that helps.
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4, 11, 18 ... 81 forms an Arithmetic Progression.

No of terms = (Last term - First term)/Common Diff + 1 = (81 - 4)/7 + 1 = 12

For more on number of terms in an AP, check: http://www.veritasprep.com/blog/2012/03 ... gressions/

or look at the patterm
4 = 4 + 0*7
11 = 4 + 1*7
18 = 4 + 2*7
.
.
.
81 = 4 + 11*7

So the terms start from 0 and go on till 11 i.e. 12 terms
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Hi,

I used the following method: it may seem long and the steps may seem complicated but most of the steps can be done in the head......below is the thought process:- Further more, in this example, it is possible to count the 12 multiples......but in case a similar problem is encountered with a large number of multiples
that is cumbersome to count...then the following method can be useful...

we are given that any term = 5 + (n-1) 3 ----> this confirms that the series is in Arithmetic Progression and the Nth term is "a+(n-1)d"
=> First term = 5 and common difference "d" = 3

we are also given that n varies from 1 to 85

therefore the series is 5,8, 11, 14......257


In the series above 14 is the 1st number which is divisible by 7. Now, because the series will proceed with increments of 3 (common difference), the only next multiple of 7 will be the number in the series when 3 will have been added 7 times or in other words, when a cumulative increment of 21 would have taken place.....=> 14, 35 , 56, 77..... (till a number less than 257)

The above series is also an AP with a common difference of 21...

Nth term of the series will be 14 + (n-1)*21 < 257
=> (n-1)< 257/13
=> (n-1)< 12
=> n<13

Now, Check for n= 12


14+ (11)21 = 245 ----> falls within 5 to 257 range.....therefore we have 12 factors....

Tn = 5+3(n-1) = 3n+2.

Let 3n+2 = 7k (k has to be an integer)

or k = (3n+2)/7. Now we find the first value of n, where n = [1,85] which gives k an integral value.

We see that n=4 gives k as 2. Thus, value of n will be 4 and values where you keep adding 7 i.e 4,11,18 etc till 85. The last value we get is n=81 and a total of 12 values for n.

E.