Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 32)
32.
How many 3-digit numbers are completely divisible 6 ?
Answer: Option
Explanation:
3-digit number divisible by 6 are: 102, 108, 114,... , 996
This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then tn = 996.
a + (n - 1)d = 996
102 + (n - 1) x 6 = 996
6 x (n - 1) = 894
(n - 1) = 149
n = 150
Number of terms = 150.
Discussion:
33 comments Page 1 of 4.
Ashok Kumar said:
1 decade ago
n=[(l-a)/d]+1
n=150
n=150
Vivek said:
1 decade ago
Adding the given three deghit it means =150
1+5+0/6=remander is 0
Hence if the adding number is not same in divedent and remander is not 0.
1+5+0/6=remander is 0
Hence if the adding number is not same in divedent and remander is not 0.
Mahesh Patil said:
1 decade ago
Total Three digit numbers = 999-100+1 = 900
so numbers Divisible by 6 = 900 / 6 = 150 (Only Quotient ... don't consider the remainder)
so numbers Divisible by 6 = 900 / 6 = 150 (Only Quotient ... don't consider the remainder)
Rajkin said:
1 decade ago
Thanks mahesh patel.
Sworna said:
1 decade ago
@Mahesh.
Why it is 999-100+1? I can't understand please help me.
Why it is 999-100+1? I can't understand please help me.
Neha nagar said:
1 decade ago
@sworna
Dear sworna here +1 is because 100 is also included in three digit no we can not leave 100 here. So it will b better to solve like this....3 digit nos-2 digit nos. So that 999-99=900. Hope you have got it.
Dear sworna here +1 is because 100 is also included in three digit no we can not leave 100 here. So it will b better to solve like this....3 digit nos-2 digit nos. So that 999-99=900. Hope you have got it.
Gis said:
1 decade ago
@sworna
999-100+1 is taken based on arithmetic progression . last 3-digit no: is 999 and first 3-digit no: is 100 . according to the equation n=(l-a)/d +1
here l=999,a=100, d=1
so n=(999-100)/1 +1=900
hope u can understand
999-100+1 is taken based on arithmetic progression . last 3-digit no: is 999 and first 3-digit no: is 100 . according to the equation n=(l-a)/d +1
here l=999,a=100, d=1
so n=(999-100)/1 +1=900
hope u can understand
Lalit said:
1 decade ago
Another short technique is last 3 digit no is 999 so we divide 999/6= 166 we don't consider decimal part. Now, we only need 3 digit number so 100/6= 16 therefore, 166-16=150 answer.
166 is the total number divisible by 6. We subtracted 100 because before 100 there is only 2 digit number divide by 6.
166 is the total number divisible by 6. We subtracted 100 because before 100 there is only 2 digit number divide by 6.
Ashish said:
1 decade ago
If we want to find how many numbers present in 3 digit no which is divisible by any number. Then here is method.
Eg: How many 3-digit numbers are completely divisible 8 ?
Answer:
1000/8=125.
100/8=12 with remainder 4.
125-12=113.
So 113 numbers are divisible from 100 to 999 by 8.
Eg: How many 3-digit numbers are completely divisible 8 ?
Answer:
1000/8=125.
100/8=12 with remainder 4.
125-12=113.
So 113 numbers are divisible from 100 to 999 by 8.
Harpreet said:
1 decade ago
@Ashish total numbers divisible by 8 comes 112 & not 113.
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