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
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6 Answers
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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