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## mathmath333 one year ago counting question

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1. mathmath333

Using the digits $$\{0,1,2,3,4,5\}$$ form a $$4$$ digit number divisible by $$4$$ without repetition.

2. ganeshie8

Look at the decimal expansion of any integer : $(\cdots abcd)_{10} = \cdots +a*10^3+a*10^2+c*10+d = 10^2*M + c*10+d$

3. ganeshie8

since 4 divides 10^2, the remainder is simply $$c*10+d$$ so you just need to check that part of the number

4. ganeshie8

In other words, if the number formed by "last two digits" is divisible by 4, then the number itself is divisible by 4

5. mathmath333

how to use this $$c*10+d$$

6. mathmath333

the question should be this Using the digits {0,1,2,3,4,5} in how many ways can form a 4 digit number divisible by 4 be formed without repetition.

7. phi

I would use brute force: list all 2 digit combos divisible by 4 cross off any that have repetitions of digits, or digits not in your set

8. phi

that will be a good start

9. ganeshie8

I would do the same, perhaps there is a smarter way to work this using symmetry... need to think it through..

10. anonymous

so we want $$a\cdot10^3+b\cdot10^2+c\cdot10+d$$ to be congruent to $$0\pmod4$$: $$a\cdot 2^3+b\cdot 2^2+c\cdot 2+d\equiv 0\pmod 4\\2c+d\equiv 0\pmod 4\\2c\equiv -d\pmod 4$$now consider that $$2$$ is not invertible $$\pmod 4$$, so look at $$c=0,1,2,3,4,5$$: $$c=0\implies d=4\\c=1\implies d=2\\c=2\implies d\in\{0,4\}\\c=3\implies d=2\\c=4\implies d=0\\c=5\implies d=2$$ in other words, the two-digit multiples of 4 that use no digits bigger than 5 and have no repeating digits we have here are: $$4,12,20,24,32,40,52$$which you could've found easily via exhaustive search (but the above just confirms it)

11. phi

the next wrinkle is I assume 4 digits means no leading 0 so you should break your 7 items into those that have a 0 (which allows the remaining 2 digits to be any of the remaining) and those without a 0

12. mathmath333

Using the digits $$\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\$$

13. anonymous

and then for the total four digit numbers, we have 3 spots; our last spot corresponds to the final two digits, giving 7 possibilities, using up two of the six possible digits and leaving 4 and 3 possibilities for the other two spots, so $$7\cdot4\cdot3=84$$. this presumes we allow for leading zeros

14. mathmath333

i will count this case $$\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\$$ by 4*4*3 ?

15. phi

for those 3 cases, you have 4 numbers left over you have 4 choices for the leading digit, and 3 for the next thus 4*3*3

16. phi

for the 4 remaining patterns, 12,24,32,52 you have 3 choices for the leading digit (assuming no leading 0) and 3 choices for the 2nd (we can use the 0) thus 3*3*4

17. phi

for no leading 0, you should get 72 if we allow leading 0's , it's 7*4*3= 84 as oldrin posted up above.

18. mathmath333

4*3*3+3*3*4 =72 , the ordirin bataku gave 84

19. mathmath333

72 is correct ?

20. ganeshie8

yeah 72 looks good to me

21. phi

It is a judgement call. If we do not allow leading zeros then 72 if they are allowed then 84 but I am assuming no leading zeros (and the questioner gets marks off for being vague)

22. mathmath333

leading 0 is not alloewd

23. mathmath333

*allowed

24. mathmath333

thnx all

25. ganeshie8

save computers, allowing leading zeroes makes the number of digits of a number undefined because we can put arbitrary number of leading zeroes before any number

26. mathmath333

save the tiger

27. ganeshie8

save trees

28. sohailiftikhar

so what if we find a probability ?

29. sohailiftikhar

by fixing 0 at the end find the numbers

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