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How many 4 digit number can be formed by 0,1,2,3,4,5 divisible by 4 with repetition

My Approach:

Last two digits can be 00,04,12,20,24,32,40,44,52 that is 9 possibilities for last two digits.

For the hundredth place digit all 6 possibilities exist.For the thousand place we have 5 options (1,2,3,4,5 ).

Hence the solution is 5*6*9=270

Is my solution correct? Please correct me if I am wrong.

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    $\begingroup$ Looks good to me. $\endgroup$ – 5xum Aug 25 '15 at 11:05
  • $\begingroup$ You should state explicitly that that digits are allowed to repeat. $\endgroup$ – Thomas Andrews Aug 25 '15 at 11:06
  • $\begingroup$ @ThomasAndrews i did $\endgroup$ – justin takro Aug 25 '15 at 11:09
  • $\begingroup$ You stated without repetition!!! $\endgroup$ – barak manos Aug 25 '15 at 11:10
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    $\begingroup$ Possible duplicate of What will be the probability in this case? $\endgroup$ – Resorcinol Jan 28 '17 at 14:01
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Your ans of $270$ is correct, but you might find this way simpler:

Since a number can't start with zero, there will be $5\cdot6^3$ valid numbers,
of which exactly $1/4$ will be divisible by $4$

[Think of a number system to base 6 to understand the rationale]

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There are 5*6*6*6 = 1080 possible numbers. Half of which, 540, is divisible by 2. Half of that again, 270, is divisible by 4.

Edit: Perhaps not obvious. But we can observe, that the numbers are consecutive, except gaps of length 4, 44, or 444; all gaps multiple of 4.

Edit2:

In a sequence of consecutive numbers every fourth number is divisible by 4. Modulo 4 the sequence goes 0, 1, 2, 3, 0, 1 ...

In the increasing sequence of numbers made from only the digits 0, 1, 2, 3, 4 and 5, there will be gaps. We can observe that the gaps will have lengths of 4, 44 or 444. All gaps are a multiple of 4.

Therefore the modulo 4 sequence must run systematically and always increase by 1 modulo 4 across a gap.

There are 5*6*6*6 = 1080 four digit numbers in the sequence. A multiple of 4, so 1/4, i.e. 270 of the numbers, are divisible by 4.

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    $\begingroup$ It's not obvious to me that this set divides exactly into quarters. I guess it's not hard to prove though. $\endgroup$ – Empy2 Aug 26 '15 at 8:08
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    $\begingroup$ This surely works as an estimate (and is what I first did for estimation) but doesn't generally work as exact answer, for example $\operatorname{card}(\{3,4\}\cup \{6,7,8\}\cup\{11,12,13\}) = 8$, but there are $3$ numbers divisible by $4$ in that set, not $8:4 = 2$. $\endgroup$ – Ennar Aug 26 '15 at 9:03
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Best way is to identify the first(1000) and last 4 digit numbers(5552) divisible by 4. Then they will be in arithmetic progression with d=4. Then you can identify n=number of terms[(5552-1000)/4+1=1139 numbers can be formed], which gives you exact answer for any such problems with repetition.

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  • $\begingroup$ "They will be in AP with d = 4". Not correct.1008, for example, is invalid, as are any numbers having digit(s) > 5. $\endgroup$ – true blue anil Jun 1 '18 at 18:56
  • $\begingroup$ You can instead use the AP with d=4 in a number system to base 6 $\endgroup$ – true blue anil Jun 2 '18 at 6:16

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