How many six digit nos. can be formed using the digits 0 to 5, without repetition such that the nos is divisible by the digit at its units place?
How many six digit nos. can be formed using the digits 0 to 5, without repetition such that the nos is divisible by the digit at its units place?
The correct option is B 426
Case 1-》0 in the units digit:
This is not possible as any number divided by zero is not defined
Case 2-》1,2,3 or 5 in the units place:
If 1,2,3, or 5 are there in the units digit,irrespective of the rest of the digits, the 6 digit number will always be divisible by the digit in the units place. This is because:
1-》every number is divisible by 1
2-》for a number to be divisible by 2 there must be an even number in the units place and 2 is an even nimber
3-》for a number to be divisible by 3,the sum of the digits should be divisible by 3 and the sum of the digits=0+1+2+3+4+5=15 which is divisible by 3
5-》for a number to be divisible by 5,there must be either 0 or 5 in the units place and this criteria is being satisfied
Now,the total possible numbers with 1,2,3, or 5 in the units place are:
Fixing 1 in the units place:
4*4*3*2*1=96
Now the first term in the equation is 4 as 0 cant be the digit as in that case it will become a five digit number.The rest of the terms are on the basis of total possible outcomes.
Now the answer will be the same for 2,3,5 in the units place
So the total possible numbers with 1,2,3,5 in the units place is:
4*96=384——-(1)
Case 3-》4 in the units place:
For a number to be divisible by 4 the last 2 digita must be divisible by 4(for eg-112 is divisible by 4 as the last 2 digits,12,is divisible by 4)
So if we fix 4 in the units place,we need to fix a number in the tenths place so that the number is divisible by 4 :
04
14
24
34
54
Out of the above,04 and 24 are divisible by 4 (44 is not possible as repetition is not allowed)
Now lets take 04 as the last 2 digits:
The total possible numbers with 04 as the last 2 digits are
4*3*2*1=24——(2)
Now the total numbers with 24 as the last 2 digits are:
3*3*2*1=18—-(3)
The first term in the equation is 3 as 0 cant be the first digit.
Now its just a matter of adding the answers that we got in (1),(2) and (3)
So the number of 6 digit numbers divisible by the number in the units place is:
384+24+18=426
Case 1-》0 in the units digit:
This is not possible as any number divided by zero is not defined
Case 2-》1,2,3 or 5 in the units place:
If 1,2,3, or 5 are there in the units digit,irrespective of the rest of the digits, the 6 digit number will always be divisible by the digit in the units place. This is because:
1-》every number is divisible by 1
2-》for a number to be divisible by 2 there must be an even number in the units place and 2 is an even nimber
3-》for a number to be divisible by 3,the sum of the digits should be divisible by 3 and the sum of the digits=0+1+2+3+4+5=15 which is divisible by 3
5-》for a number to be divisible by 5,there must be either 0 or 5 in the units place and this criteria is being satisfied
Now,the total possible numbers with 1,2,3, or 5 in the units place are:
Fixing 1 in the units place:
4*4*3*2*1=96
Now the first term in the equation is 4 as 0 cant be the digit as in that case it will become a five digit number.The rest of the terms are on the basis of total possible outcomes.
Now the answer will be the same for 2,3,5 in the units place
So the total possible numbers with 1,2,3,5 in the units place is:
4*96=384——-(1)
Case 3-》4 in the units place:
For a number to be divisible by 4 the last 2 digita must be divisible by 4(for eg-112 is divisible by 4 as the last 2 digits,12,is divisible by 4)
So if we fix 4 in the units place,we need to fix a number in the tenths place so that the number is divisible by 4 :
04
14
24
34
54
Out of the above,04 and 24 are divisible by 4 (44 is not possible as repetition is not allowed)
Now lets take 04 as the last 2 digits:
The total possible numbers with 04 as the last 2 digits are
4*3*2*1=24——(2)
Now the total numbers with 24 as the last 2 digits are:
3*3*2*1=18—-(3)
The first term in the equation is 3 as 0 cant be the first digit.
Now its just a matter of adding the answers that we got in (1),(2) and (3)
So the number of 6 digit numbers divisible by the number in the units place is:
384+24+18=426
