# 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

neela | High School Teacher | (Level 3) Valedictorian

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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 ).

william1941 | College Teacher | (Level 3) Valedictorian

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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.