Aptitude - Numbers - Discussion

The number between 1 to 100 number which a are divisible by 6 is;

6 (100(16
___
96
___
4.

Hence, the number which is divisible between 1 to 100 by 6 are "16".

Now we will find the number which is divisible between 1 to 23.

6)23(3
18
_
5.

Hence the number which is divisible between 1 and 23 are "3".
Now the number between (1 to 100 ) - ( 1 to 23).
16 - 3 = 13 => Answer.

How about this one?

100-23 = 77 divided it by 6 77/6 = 12.83 round off 13.

It was a difficult one to remember the numbers which are divisible by 6.

So, I recommended you to solve this by the following method.

Formula: n=(l-a)/d+1.
For the above question l=100,a=23, d=6.
n=((100-23)/6)+1.
=(77/6)+1.
=77+6/6.
= 83/6 = 13 (consider only the quotient, not the remainder).

@Ramskittu.

I am not understanding your method for getting solution. Please explain it again.

@Shiv here is the method for you.

Here total 78 numbers are here, check by AP formula.

Tn = a+(n-1) d where a = 23, d =1, Tn = 100, so n = 78. Now simply divide 78/6 = 13.

How can we know which number is perfectly divisible by 78?

9944 or 9988.

Can anyone sole it?

Solution :

N = (101-23)/6,
N = 78/6,
N = 13.

Reason for subtraction of 101 from 23 is that we need all the number from 23 to 100 including 100. If we subtract 100-23 we are neglecting 100. For eg 10-1 gives 9 numbers neglecting 10.

Now why dividing by 6? Division means to group something in parts. So while dividing by 6 we group each numbers in pair of 6 out of which one number is divisible by 6.

Therefore as many number of groups = as many numbers which are divisible by 6

Total numbers between 23 to 100 = 100-23+1 = 78

Numbers Divisible by 6 = 78 / 6 =13 (consider Only Quotient ....not Remainder)

24+(n-1)6=96
=n-1=12
=n=13

4*6 = 24
5*6 = 30
6*6 = 36
7*6 = 42
8*6 = 48
9*6 = 54
10*6 = 60
11*6 = 66
12*6 = 72
13*6 = 78
14*6 = 84
15*6 = 90
16*6 = 96
TOTAL = 13.