Aptitude - Numbers - Discussion

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).

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

9944 or 9988.

Can anyone sole it?

Thanks for explaining this.

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.

@Ramskittu.

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

For example,

Number b/w 15 to 180 divided by 5 are,

180-15+1 = 166.

166/5 = 33.xxx.

Then 33 is the right answer?

How about this one?

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

Max number which or completely divisible by 6 below 100 is 96.

So 96/6 = 16.

Remove before divisible numbers of 23. i.e. 3.

So finally 16-3 = 13.

@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.

Any other method please.