counting question

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Using the digits \(\{0,1,2,3,4,5\}\) form a \(4\) digit number divisible by \(4\) without repetition.
Look at the decimal expansion of any integer : \[(\cdots abcd)_{10} = \cdots +a*10^3+a*10^2+c*10+d = 10^2*M + c*10+d\]
since 4 divides 10^2, the remainder is simply \(c*10+d\) so you just need to check that part of the number

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In other words, if the number formed by "last two digits" is divisible by 4, then the number itself is divisible by 4
how to use this \(c*10+d\)
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.
  • 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
  • phi
that will be a good start
I would do the same, perhaps there is a smarter way to work this using symmetry... need to think it through..
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)
  • 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
Using the digits \(\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\ \)
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
i will count this case \(\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\\) by 4*4*3 ?
  • 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
  • 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
  • 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.
4*3*3+3*3*4 =72 , the ordirin bataku gave 84
72 is correct ?
yeah 72 looks good to me
  • 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)
leading 0 is not alloewd
*allowed
thnx all
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
save the tiger
save trees
so what if we find a probability ?
by fixing 0 at the end find the numbers

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