a Sequence is F(n)=[5-(n-1)]/3, n=1,...,85 integers how many numbers in this sequence are divisible by 7?
Answer is 3 for this one!
sequences tough problem
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Anurag@Gurome
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F(n) = 5 - 3(n - 1) = 8 - 3ngvosough wrote:a Sequence is F(n)=5-(n-1)*3, n=1,...,85 integers how many numbers in this sequence are divisible by 7?
Hence, the sequence is 5, 2, -1..., -247
Hence, if there is any number in the sequence which is divisible by 7, that must be a negative integer.
Now, if F(n) is divisible by 7 then F(n) must be of the form -7k, where k is some positive integer. As minimum value of F(n) is -247, k must be less than 247/7 = 35.(Something)
So, 8 - 3n = -7k
--> 3n = 7k + 8 = (6k + 6) + (k + 2) = 3*(2k + 2) + (k + 2)
Hence, (k + 2) must be a multiple of 3.
Hence, possible values of k : 1, 4, 7..., 34
Hence, number of possible values of k = (34 - 1)/3 + 1 = 12
Hence, there are 12 numbers in this sequence which are divisible by 7.
Note : We can also also find the terms of the sequence which will be of the form -7k but that's not necessary. They are : -7, -28, -49, ..., -238.
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nandy1984
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F(n) = 5 - (n-1)*3 , n=1,.....85gvosough wrote:a Sequence is F(n)=5-(n-1)*3, n=1,...,85 integers how many numbers in this sequence are divisible by 7?
Thanks..
Lets simplify this
F(n) = 8 - 3n
This should be divisible by 7
n------------->>>>>>8-3n---------------------->>>>8-3n divisible by 7?
1------------->>>>>>5------------------------------->>>>NO
2------------->>>>>>2------------------------------->>>>NO
3------------->>>>>>-1------------------------------>>>>NO
4------------->>>>>>-4------------------------------>>>>NO
Once you know the sequence no need to calculate always just reduce 3 from the above value of 8-3n
5------------->>>>>>-7------------------------------>>>>YES
6------------->>>>>>-10------------------------------>>>>NO
7------------->>>>>>-13 ----------------------------->>>>NO
8------------->>>>>>-16------------------------------>>>>NO
9------------->>>>>>-19------------------------------>>>>NO
10------------->>>>>>-22------------------------------>>>>NO
11------------->>>>>>-25------------------------------>>>>NO
12------------->>>>>>-28------------------------------>>>>YES
Now we can observe that its following some pattern of Arithematic Progresssion
When n=5,12 its divisible by 7 now we can predict the numbers when the F(n) will be divisible like n = 5,12,19,26.....
Now how to find the number of numbers???
In an arithematic progression T(n) = a + (n-1)*d, where "a" is the first term, 'd' is the increment or decrement value, n is the nth term. Our final number is 85.
So here a = 5 from where the equation ins divisible by 7, d = 7 ( 5+7 =12, 12+7 = 19.....)
n= ?, T(n) = 85.
85 = 5 + (n-1) * 7
80 = (n-1) * 7
(n-1) ~ 11
n = 11+1 = 12
So there are total 12 terms
If you are confused you can manually add like this
n = 5,12,19,26,33,40,47,54,61,68,75,82. So here there are 12 terms....
Hope i am clear...Sorry for making it so BIG... The answer seems leangthy but if u start doing it you will solve it in less than 1.5mins....Tc
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saketk
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gvosough wrote:a Sequence is F(n)=5-(n-1)*3, n=1,...,85 integers how many numbers in this sequence are divisible by 7?
Thanks..
F(n) = 8- 3n
1 -- 5
2-- 2
3 -- -1
4 -4
5 -7 ------- every 7th number from here on will be divisible by 7
12 -28 ----
19 -49 etc...
It's an Ap (arithmetic series) with the common difference = (-3)
next -- 26, 33, 40, 47, 54, 61, 68, 75, 82 [ the values of n] we can find the corresponding values of f(n)
Total 12
