Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 33)
33.
How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?
Answer: Option
Explanation:
Required numbers are 24, 30, 36, 42, ..., 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then tn = 96 a + (n - 1)d = 96
24 + (n - 1) x 6 = 96
(n - 1) x 6 = 72
(n - 1) = 12
n = 13
Required number of numbers = 13.
Discussion:
16 comments Page 1 of 2.
Pratyush Sharma said:
5 months ago
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
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
(1)
Crr said:
1 year ago
Thanks for the explanation.
Bhuvana ram said:
6 years ago
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).
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).
(3)
Sarbarth said:
8 years ago
How can we know which number is perfectly divisible by 78?
9944 or 9988.
Can anyone sole it?
9944 or 9988.
Can anyone sole it?
(1)
ABDUL HASIB LASKAR said:
8 years ago
Thanks for explaining this.
Haseeb said:
9 years ago
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.
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.
(14)
Vijay said:
9 years ago
@Ramskittu.
I am not understanding your method for getting solution. Please explain it again.
I am not understanding your method for getting solution. Please explain it again.
(2)
Rehana said:
9 years ago
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?
Number b/w 15 to 180 divided by 5 are,
180-15+1 = 166.
166/5 = 33.xxx.
Then 33 is the right answer?
Nano_Desu said:
10 years ago
How about this one?
100-23 = 77 divided it by 6 77/6 = 12.83 round off 13.
100-23 = 77 divided it by 6 77/6 = 12.83 round off 13.
(5)
Ramskittu said:
1 decade ago
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.
So 96/6 = 16.
Remove before divisible numbers of 23. i.e. 3.
So finally 16-3 = 13.
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