How many 5 digit numbers that are divisible by 3 can be formed using the digits 0?

How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?

Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition.
For a number to be divisible by 3, the sum of its digits should be divisible by 3,
Consider the digits: 1, 2, 3, 4 and 5
Sum of the digits = 1 + 2 + 3 + 4 + 5  = 15
15 is divisible by 3.
∴ Any 5-digit number formed using the digits 1, 2, 1
Starting with the most significant digit, 5 digits are available for this place.
Since, repetition is not allowed, for the next significant place, 4 digits are available.
Similarly, all the places can be filled as:

Number of 5-digit numbers
= 5 × 4 × 3 × 2 × 1 = 120
Now, consider the digits: 0, 1, 2, 4 and 5
Sum of the digits = 0 + 1 + 2 + 4 + 5 = 12 which is divisible by 3.
∴ Any 5-digit number formed using the digits
0, 1, 2, 4, and 5 will be divisible by 3.
Starting with the most significant digit, 4 digits are available for this place (since 0 cannot be used).
Since, repetition is not allowed, for the next significant place, 4 digits are available (since 0 can now be used).
Similarly, all the places can be filled as:

Number of 5-digit numbers
= 4 × 4 × 3 × 2 × 1 = 96
Next, consider the digits: 0, 1, 2, 3, 4
Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3.
∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3.
Further, no other selection of 5 digits (out of the given 6)
will give a 5-digit number, which is divisible by 3.
∴ Total number of 5adigit numbers divisible by 3
= 120 + 96 = 216

• Forum
• Algebra
• Permutation and Combination Question

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6 Answers

Swapnil Saxena

102 Points

10 years ago

APURV GOEL

39 Points

10 years ago

I know the answer but i want the solution that takes less time because i did it by cases method  which was vey time consuming.

nikhil yadala

38 Points

10 years ago

hi

i have got the answer 336

if the answer is correct,you approve this .then i will write the solution also

nikhil yadala

38 Points

10 years ago

we have to get 5 digit numberrs which are divisible by 3

the divisibility test of 3 is that the sum of the digits of that number would be divisible by 3

from the given digits we can hav 3 sets of numbers which are divisible by 3

they are  all the 5 digit numbers formed by the digits (2,3,4,7,8)  we get 5 factorial number of 5 didit numbers ie 120
                                                                           (2,3,7,8,1)   we get 5 factorial number of 5 didit numbers ie 120
           (7,8,2,1,0)    we get 4*4*3*2*1  5 digit numbers =96 

totally 120+120+96 =336

hence the answer is 336

G Narayana Raju

51 Points

10 years ago

we should see the no of combinations

without zero

7,8,3,2,1 make 5!

2,3,4,7,8 make 5!

8,3,4,2,1 make 5!

1,2,4,8,3 make 5!

with zero

7,4,3,1,0 makes 4x4!

8,4,1,2,0 makes 4x4!

7,3,2,1,0makes 4x4!

so we get the no of combinations of five digit no which are divisible by three.

I am not sure if I wrote all the combinations.

nikhil arora

54 Points

10 years ago

the sum of all given no.s is = 0+1+2+3+4+7+8 = 25 (also this is the maximum sum of digits that can be obtained)

a no will be divisible by 3 if its sum is 24 or 21 or 18 ... so on.

 also we have to select 5 digits from 7 digits we have to omit 2 digits

so for a sum of 24 , we have to omit those no which have a sum of 1 i.e.------ 0,1

similarly

for 21 ----we have to omit those which have a sum of 4------ 0,4 or 1,3

for 18 ----- 0,7 or 4,3

for  15 ----- 2,8 or 7,3

for 12 ----  no combination is possible.

we have 7 cases in all.

for three cases the 5 digit no. will not contain 0 so no. of ways ----- 3 *(5!) = 3*120 = 360

for rest four cases , the 5 digit no. will contain 0 hence no of ways.----- 4 * (4x4x3x2x1)= 384

total no of ways = 384 + 360= 744

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