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mathmath333
 one year ago
counting question
mathmath333
 one year ago
counting question

This Question is Closed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Using the digits \(\{0,1,2,3,4,5\}\) form a \(4\) digit number divisible by \(4\) without repetition.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Look at the decimal expansion of any integer : \[(\cdots abcd)_{10} = \cdots +a*10^3+a*10^2+c*10+d = 10^2*M + c*10+d\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3since 4 divides 10^2, the remainder is simply \(c*10+d\) so you just need to check that part of the number

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3In other words, if the number formed by "last two digits" is divisible by 4, then the number itself is divisible by 4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how to use this \(c*10+d\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I would do the same, perhaps there is a smarter way to work this using symmetry... need to think it through..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 twodigit 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
 one year ago
Best ResponseYou've already chosen the best response.2the 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

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Using the digits \(\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\ \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i will count this case \(\text{_._} 04 \\ \text{_._} 20 \\ \text{_._} 40 \\\) by 4*4*3 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2for no leading 0, you should get 72 if we allow leading 0's , it's 7*4*3= 84 as oldrin posted up above.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.14*3*3+3*3*4 =72 , the ordirin bataku gave 84

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah 72 looks good to me

phi
 one year ago
Best ResponseYou've already chosen the best response.2It 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)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1leading 0 is not alloewd

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3save 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

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.0so what if we find a probability ?

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.0by fixing 0 at the end find the numbers
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