How many 5 digit numbers are divisible by 4 formed using 0 1 2 4 6 8?
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If abcdeabcde is the general form of a five digit number, where a,b,c,d,e stand for the digits, then it is divisible by 3 ⟺⟺ a+b+c+d+ea+b+c+d+e is divisible by 3.
Checking for those numbers. 5+4+3+2+1=155+4+3+2+1=15 which is divisible by 3. So any permutation of these 5 numbers will give all five digit numbers formed by {1,2,3,4,51,2,3,4,5} divisible by 3
And that number is 5∙4∙3∙2∙1=1205•4•3•2•1=120
Now the only digit we omitted is 0. So if I replace any digit of the five digit number abcdeabcde, then I should remove one digit also such that the sum of the digit is still a multiple of three. Since I'm replacing a digit by 0, the sum is not yet changed. So I have to take out a multiple of 3 to make the sum still divisible by 3. We have only one multiple of 3 in {1,2,3,4,51,2,3,4,5} which is 3 itself. So we remove it. Now the set is {0,1,2,4,50,1,2,4,5}
And the numbers in our discussion are those formed by {0,1,2,4,50,1,2,4,5}
Note that 5+4+2+1=125+4+2+1=12 still a multiple of three as we expected.
Total 5 digit numbers without any digit repeated, formed by this set is 4∙4∙3∙2∙1=964•4•3•2•1=96
All the possible combinations come under the above 120 and 96 numbers.
So the total 5 digit numbers formed by
{0,1,2,3,4,50,1,2,3,4,5} without repetition divisible by 3 is 120+96=216120+96=216 numbers.
Repetition not allowed 216216 numbers.
Checking for those numbers. 5+4+3+2+1=155+4+3+2+1=15 which is divisible by 3. So any permutation of these 5 numbers will give all five digit numbers formed by {1,2,3,4,51,2,3,4,5} divisible by 3
And that number is 5∙4∙3∙2∙1=1205•4•3•2•1=120
Now the only digit we omitted is 0. So if I replace any digit of the five digit number abcdeabcde, then I should remove one digit also such that the sum of the digit is still a multiple of three. Since I'm replacing a digit by 0, the sum is not yet changed. So I have to take out a multiple of 3 to make the sum still divisible by 3. We have only one multiple of 3 in {1,2,3,4,51,2,3,4,5} which is 3 itself. So we remove it. Now the set is {0,1,2,4,50,1,2,4,5}
And the numbers in our discussion are those formed by {0,1,2,4,50,1,2,4,5}
Note that 5+4+2+1=125+4+2+1=12 still a multiple of three as we expected.
Total 5 digit numbers without any digit repeated, formed by this set is 4∙4∙3∙2∙1=964•4•3•2•1=96
All the possible combinations come under the above 120 and 96 numbers.
So the total 5 digit numbers formed by
{0,1,2,3,4,50,1,2,3,4,5} without repetition divisible by 3 is 120+96=216120+96=216 numbers.
Repetition not allowed 216216 numbers.
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