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Consider a sequence of numbers given by the expression 5 + (

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Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 06 Dec 2012, 05:28
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Oct 2013, 23:47, edited 3 times in total.
Edited the question and added the OA.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 06 Dec 2012, 05:51
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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 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.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 06 Dec 2012, 05:57
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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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 06 Dec 2012, 06:06
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hii Bunuel.
This is rather the same approach used by me.
Is there any alternative approach to the solution?
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 07 Dec 2012, 20:26
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.

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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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 08 Dec 2012, 04:46
Marcab wrote:
hii Bunuel.
This is rather the same approach used by me.
Is there any alternative approach to the solution?


Hi Marcab,

How do you get this n value as brunuel stated
when n is 4, 11, 18, ..., 81 ?
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 08 Dec 2012, 09:45
Expert's post
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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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 08 Dec 2012, 11:59
Marcab wrote:
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


Thanks marcab..
Ya it was silly mistake...
i didnt recognize difference 4-11-18...
hmm this is where i loss score...

But sorry marcab i didnt get ur experiment in this question?
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 08 Dec 2012, 20:00
Expert's post
shanmugamgsn wrote:
Marcab wrote:
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


Thanks marcab..
Ya it was silly mistake...
i didnt recognize difference 4-11-18...
hmm this is where i loss score...

But sorry marcab i didnt get ur experiment in this question?


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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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 18 Dec 2012, 12:24
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the first term of the sequence is 5
the last term = 5+(85-1)*3=257
the number of multiples of 7 in the range between 5 and 257 = (last multiple of 7 - first multiple of 7)/7+1 = (252-7)/2+1=36
That means that between 5 and 257 there are 36 inregers that are divisible by 7
In the given sequence, the difference between the terms is 3. That means that the sequence contains not every integer but every third integer between 5 and 257. Hence the number of multiples of 7 in the given sequence = 36/3=12
Answer E
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 18 Dec 2012, 23:04
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Amateur wrote:

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.

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


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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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 19 Dec 2012, 06:06
The equation has to be divisible by multiples of 7:

Hence, it should equate to equate to 7, 14, 21......77. Which will give us 12 values for n.

Hence, Option E.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 19 Dec 2012, 07:30
Marcab wrote:
hii Bunuel.
This is rather the same approach used by me.
Is there any alternative approach to the solution?


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....
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 27 Jan 2013, 19:26
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Responding to a pm:

HI Karishma
I need to clarify a logical doubt on the question.
If the value of n is put in the formula and sequence of number is obtained, we get the following
5,8,11,14,17,...................................,245,248,251,254,257

Formula that i know to find out the number of terms in a sequence (A.P.) or more precisely number of multiples of x in a sequence (A.P.) is
[(Last multiple of x in the range - first multiple of x in the range)/x] + 1
hence
Last multiple of 7 in the sequence is 245
First multiple of 7 in the sequence is 14
Hence total multiple of 7 in the sequence is [(245- 14)/7]+1 = 34
Its not even mentioned in the answer choices ...i know there is a basic fundamental error in my approach......

Analogy to above question which i used was:-
Question:- Find the total number of multiples of 4 from 20 to 99?
I use the same formula as above
[(96-20)/4]+1
=20 and it is the correct answer.....
Is there any difference between the two questions...both are A.P sequence....
Pls help me in getting my fundamental correct....


Important: It is good to know formulas since they help us save time but you must know the exact inputs and limitations of every formula you would like to use.

The formula you used is absolutely fine and very useful actually. This is what happens in the 2nd question:

Question:- Find the total number of multiples of 4 from 20 to 99.
The given sequence is 20, 21, 22, 23, 24, 25, 26, 27, 28, ... 96, 97, 98, 99
You need to find the multiples of 4 in this sequence which are 20, 24, 28, ... 96 (this is an AP)
20 is the first multiple and 96 is the last. The common difference between two consecutive multiples is 4. So you use the AP formula as you have done. Great.

The original question is a little different.
5 + (n - 1) * 3=3n+2
Putting values for n, the sequence we get is this: 5,8,11,14,17, 20, 23, 26, 29, 32, 35, 38, ...................................,245,248,251,254,257
We need to find the multiples of 7 in this sequence which are 14, 35, ... 245 (mind you, we need the multiples of 7 which appear in this sequence - 21, 28 do not appear in this sequence)

The first multiple is 14 and the last is 245 and the common difference is 21.
Now use the formula: (245 - 14)/21 + 1 = 12 and that's your answer!
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 30 Jan 2013, 06:54
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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.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 31 Oct 2013, 22:02
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.

Answer: E.

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


Hi Bunuel,

Do we need to calculate all 12 numbers multiple of 7 for 3n+2? How can we generalise and find the multiples? I could not figure it out. :(
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 07 Nov 2013, 00:32
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saurabhprashar 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.

Answer: E.

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


Hi Bunuel,

Do we need to calculate all 12 numbers multiple of 7 for 3n+2? How can we generalise and find the multiples? I could not figure it out. :(


I am guessing your question is how do we find these values: 4, 11, 18 etc

3n + 2 needs to be divisible by 7. Find out the first value of n for which 3n+2 is divisible by 7. Try n = 1/2/3/4. You get that when n = 4, 3n+2 = 14 and hence divisible by 7. Now you need to find the next value of n for which 3n+2 will be divisible by 7. Should you check for n = 5/6/7/8... etc. No! It will take you forever to figure out the required values of n since n can take any value up till 85. The use of logic will help you here:
3*4 + 2 is divisible by 7. The value of n is 4 here. What will be the next value of n? It will be 4+7. The one after that will be 4+14. Next will be 4+21 and so on...
Why?
Say after 4, the next value of n that satisfies this condition is m more than 4. So we are saying that 3*(4+m) + 2 is divisible by 7.
i.e. (3*4+ 2) + 3m is divisible by 7.
i.e. 14 + 3m is divisible by 7.
Now 14 is already divisible by 7. For 3m to be divisible by 7, m must be a multiple of 7. So m could be 7, 14, 21 ... etc.

Hence n could be 4, 4+7, 4+14, 4+21, .... till 4 + 77.
These are 12 terms.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 21 Jan 2014, 05:01
Or instead of counting one can just set up an inequality such as

1<=4+7k<=85
0<=k<=11

Total of 12 terms

Hope it helps
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 21 May 2014, 23:19
The expression is Tn=5+3(n-1), the first term is 8 and last term is 257

if Tn needs to be divisible by 7 then it has to be a multiple of 7
therefore Tn=7x=5+3(n-1) [for some integer x]
alternatively, 3(n-1)=7x-5 (means we need to find out there are how many multiples of 3 for 7x-5)
if x=1, 7x-5=2, not divisible by 3
x=2, 7x-5=9, divisible by 3
x=3, 7x-5=16, not divisible by 3
x=4, 7x-5=23, not divisible by 3
x=5, 7x-5=30, divisible by 3
..........
x=8, 7x-5=51
...........
the last term before 257 for 7x-5 is 245 which satisfies 3(n-1)=7x-5

now, 9,30,51...240 forms an AP series

by applying the AP formula for Xn=X1+d(n-1)
where Xn=240, X1=9, d=21, n can be calculated as 12

the correct answer is 12

Last edited by samikpal01 on 28 May 2014, 22:51, edited 1 time in total.
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Re: Consider a sequence of numbers given by the expression 5 + ( [#permalink] New post 22 May 2014, 01:52
Did in this way:

The range given is 1 to 85, so nos divisible by 7 are

7........... 14.......... 21....... 28 ................. upto 84

Total numbers which are divisible by 7 is 12

\frac{84}{7} = 12

The equation given is

5 + (n - 1) * 3

= 5 + 3n - 3

= 3n + 2

Placing value of n = 1 to n = 85, is just changing the scale limits

from 4 (7-3) to 81 (84-3)

The numbers divisible by 7 remains the same = 12

Answer = E
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Re: Consider a sequence of numbers given by the expression 5 + (   [#permalink] 22 May 2014, 01:52
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