Aptitude - Numbers - Discussion

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.

6 written as 2*3.

The number completely divisible by both numbers 2&3.

If the number is divisible by 2 that number should be an even.

If the number is divisible by 3 then taken sum of digits.

Option verification:

Only two even numbers are there 150 & 166.

These two numbers are divisible by 2.

1+5+0 = 6(which is divisible by 3), 1+6+6 = 13(which is not divisible by 3).

Hence 150 is the answer.

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.

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.

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.

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.

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.

@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

@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.

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.