Anonymous

How Many Four-digit Numbers Can Be Formed Using The Digits 1, 2, 4, 6, 7 And 9 Such That Each Number Is Divisible By 3 But Not By 9? (Repetition Of Digits Is Not Allowed)

3

3 Answers

Oddman Profile
Oddman answered
The digits chosen must sum to a multiple of 3, but not to a multiple of 9. If no repeated digits are allowed, the combinations of digits that have the appropriate sums are
  {1, 4, 7, 9}, {2, 4, 6, 9}, {2, 6, 7, 9}
These each can be arranged in 4!=24 ways, to give a total of 3*24 = 72 unique numbers with no repeated digits.

If digits are allowed to be repeated, there are 28 choices. When digits are repeated, the number of possible variations in the digit sequence is reduced. The choices are
{1, 1, 1, 9}, {1, 1, 2, 2}, {1, 1, 4, 6}, {1, 1, 4, 9}, {1, 1, 6, 7}, {1, 2, 2, 7},
{1, 2, 6, 6}, {1, 2, 9, 9}, {1, 4, 4, 6}, {1, 4, 7, 9}, {1, 6, 7, 7}, {1, 7, 7, 9},
{2, 2, 2, 6}, {2, 2, 2, 9}, {2, 2, 4, 4}, {2, 2, 4, 7}, {2, 4, 6, 9}, {2, 4, 9, 9},
{2, 6, 6, 7}, {2, 6, 7, 9}, {4, 4, 4, 9}, {4, 4, 6, 7}, {4, 4, 7, 9}, {4, 6, 7, 7},
{6, 6, 6, 6}, {6, 6, 9, 9}, {6, 9, 9, 9}, {7, 7, 7, 9}

Altogether, there are 295 different numbers that can be made with these sets of digits.
Vin Ginge Profile
Vin Ginge answered
This answer will provide the 4-digit combinations of numbers.The actual permutations will then need constructing.
A number that is divisible by 3 has the sum of its digits equal to 3 or a multiple of 3.
As 6 and 9 are divisible by 3 then the number of 4-digit number formations are :-
1) 6 & 9 with two of the 4 remaining digits (1,2,4 and 7). The combinations of 2 of the 4 remaining digits that total 3 or a multiple of 3 are, 1&2 = 3, 2&4 = 6 and 2&7 = 9.
6+9+1+2 = 18 . This combination, although divisible by 3 is also divisible by 9 and can thus be disregarded. 6+9+2+4 = 21 which is divisible by 3 only. 6+9+2+7 = 24 which is divisible by 3 only.
2) 6 or 9 with 3 of the 4 remaining digits (exclude 9 and 6 respectively). There is only 1 combination of 3 of the remaining 4 digits that total 3 or a multiple of 3. This is 1&4&7 = 12.
6+1+4+7 = 18 This combination, although divisible by 3 is also divisible by 9 and can thus be disregarded. 9+1+4+7 = 21 which is divisible by 3 only.
3) The 4 remaining digits. As 1 + 2 + 4 + 7 = 14 is not a multiple of 3 then this combination can be disregarded.
The three successful combinations are therefore 6924, 6927 and 9147.
Within each group of numbers, the digits can be arranged 24 different ways. There are thus 3 x 24 = 72 numbers that are divisible by 3 but not by 9.

Answer Question

Anonymous