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:
34 comments Page 4 of 4.
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.
Last digit - first digit(996-102) = 894.
Divide by divisible no. 894/6 = 149.
In formula opposite (n-1) +1.
(149+1) = 150.
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.
A no. is divisible by 6 if the no. is even and sum of digit is divisible by 3.
Joshua said:
1 decade ago
What if you try the formula for geometric sequence which is an = a1 x r^n-1.
Vicky kumar verma said:
1 decade ago
999/6 = 166 ignore remainder.
99/6 = 16 ignore remainder.
Now, 166-16 = 150 answer.
99/6 = 16 ignore remainder.
Now, 166-16 = 150 answer.
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