Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 32)
32.
How many 3-digit numbers are completely divisible 6 ?
149
150
151
166
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 2 of 4.

Damini said:   8 years ago
Write numbers from sequence 149 to 387 like 149150159.

How many time does 1 occur?

Tanu said:   8 years ago
Greatest 3 digit number that is divisible by 6 is 996 (q=166).

The smallest is 102(q=17).
Apply simple number rule to count number of values between 166 to 17,
i.e. 166-17+1=150.

Thus this gives you the number of 3 digits divisible by 6.

Shradha khirid said:   8 years ago
If no is divisible by 3 and 2 then it is easily divisible by 6.
(1)

Rohit said:   9 years ago
100 - 999 these are three digits no,
(last term - first term) + 1 = 900.

And divide 900 by 6 i. e. we obtained 150 is final answer.

Samir Das said:   9 years ago
Thanks for the answer. It's useful to me.

Rahul said:   9 years ago
Please give me a solution for "What is the sum of 3 digits number which is completely divisible by 3, then find the sum".

Vicky kumar verma said:   10 years ago
999/6 = 166 ignore remainder.

99/6 = 16 ignore remainder.

Now, 166-16 = 150 answer.

Joshua said:   1 decade ago
What if you try the formula for geometric sequence which is an = a1 x r^n-1.

Ayushi said:   1 decade ago
Divisibility rule of 6.

A no. is divisible by 6 if the no. is even and sum of digit is divisible by 3.

Samir said:   1 decade ago
It can take any no.of digit so,

Last digit - first digit(996-102) = 894.

Divide by divisible no. 894/6 = 149.

In formula opposite (n-1) +1.
(149+1) = 150.


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