Aptitude - Numbers - Discussion

149 is not divisible by 6.
150 is divisible by 6.
151 is not divisible by 6.
166 is not divisible by 6.

So, the right answer is 150.

My idea is;

1+4+9=14.

1+5+0=6.

1+5+1=7.

1+6+6=13.

So, according to me, here 6 only can be purely divisible by 6.

Let the highest 3 digit num divided by 6 is 996,
then 996/6= 166.

Let the highest 2 digit num divided by 6 is 96,
then 96/6= 16,
Now 166-16= 150.

As per the Divisibility rule; If last digit is even it will divided by 2 and 3 also divided by 6. In this problem only 166 can sacrifice this method. So the answer is 166.

Lowest 3 digit number which is divisible by 3 is 102.
Highest 3 digit number which is divisible by 3 is 999.
So as per the formula a+(n-1)d=l.
The answer will be 300.

6 = 2 * 3 which are co-actors, then the number divisible by these two will also divisible by 6.

By taking the multiples of 2*3 for 6 we can operate the divisibility rule for this question.

As for 2, the ending digit should be 0, 2, 4, 6, 8 and for 3 sum of all the digit should be divisible by 3.

So, 150 is divisible by both 2 & 3. This is the correct answer.

Step 1) Consider highest 3 digit number divisible by 3 which is 996, dividing it by 6 we get remainder as 166. Since we want only 3 digit numbers which are divisible by 3 we need to subtract the 2 digit numbers in here.

Step 2) Consider highest 2 digit number divisible by 3 which is 96,dividing it by 6 we get remainder as 16.That means there are 16 (2 digit numbers) divisible by 6.

Step 3) Now subtract these two corresponding remainders we get 166-16 = 150.

Divisiablity shortcut for 6 = the number should divide be both 2 & 3.
a. 149 / 2 = no because unit digit is odd.

b. 150 /2 = yes because unit digit ends in 0, then check for 3 -> 150 / 3 = 1+5+0 = 6 = 6/3 = 2, so answer is B.

Number is divided by 6 only when it is perfectly divided by 2 and 3 both.

So Only 150 is divided by both 2 and 3.