# Find how many distinct numbers greater than 5000 and divisible by 3 can be formed from 3,4,5,6,0, each digit being used at MOST once in any number

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### 2 Answers

We are interested only the numbers generated from 3,4,5,6 and 0 using at most once. The generated number should be divisible by 3..

We know that a number is divisible 3 if the sum of all its digits is divisible by 3.

There are broadly 2 categories - 4digit number and 5 digit numbers.

(I) 4 digit numbers should have 5 or 6 in the thousandth place to exceed 5000.

a)

5 in 1000 's place: Then the other 3 digits could (0,4, 6) combination or (0 4,3 ) combination or (3,4 ,6) combinatio. So this type numbers is possible in 3!*3 = 18 ways.

b)

6 in thousand's place: After 6 we can have the digis of combinations (0,4,5) or (3,4 ,5) each with 3! number of diffrent numbers - total 3!*2 = 6*2 = 12.

II) numbers with 5 digits are all greater than 5000.

Here in the ten thousand's place we can have 4 choices from 3,4,5,6 (not zero). 1000's place could be occupied by any of the remaing other 4 digits. Similarly the 100's place could be occupied by any remaining 3, ten's place by any remaing 2 and unit's by the ultimate left out 1. So the number of different numbers = 4*4*3*2*1 = 96. Each of these numbers are divisible by 3 as the digit totals is 3+4+5+6+0 = 18 in all cases.

Adding case wise we get 18+12+96 = 126 . So 126 is the number of differerent numbers that are greater than 5000 and divisible by 3( generated by the given digits ).

We need to find the number of numbers we can form greater than 5000 and divisible by 3.

Now we see that 3+4+5+6 = 18 which is divisible by 3 so any number which has all the digits will be divisible by 3.

Therefore we can have any 5 digit number. The number of these possible is 4*4*3*2*1 = 72

Now to find the number of 4 digit numbers that satisfy this condition. We see that the only sets which are divisible by 3 are 4,5,6,3 and 5,4,3,0 and 5,4,6,0

The first digit can only be 5 or 6.

Now the numbers starting with 5 and divisible by 3 are: 5604, 5640 , 5460 , 5406, 5064, 5064, 5304, 5340, 5034, 5043, 5403, 5430, 5643, 5634, 5436, 5463, 5346, 5364. We have 18 numbers

The numbers starting with 6 and divisible by 3 are: 6540, 6504, 6450, 6405, 6045, 6054, 6345, 6354, 6435, 6453, 6534, 6543. we have 12 numbers.

Therefore the total number of numbers is 18+ 12 +72 = 102

**The required result is 102.**