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# How many five digits numbers divisible by 3 can be formed using the digits 0, l,2,3,4,7and8 if each digit is to be used atmost once.Please give me a simpler approach to solve this question.

## 6 Answers

9 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.

9 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

9 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.

9 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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