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A six digit number K is comprised of digits

I said let:

$$R=ABCDEF$$

Since $R$ is divisible by $360$ then it is also divisible by $2$. So $F=0,2,8,k$ only if $k$ is even.

$R$ is divisible by $5$ so the last digit $F=0/5$. So $F=0$

Also $R$ is divisible by $4$ so $EF$= divisible by $4$. but $F=0$ so, $EF=20,80$

Since it is divisible by $8$ then the last three digits form a number divisible by $8$. So $DEF=320, 280$

Now $ABC$ is also divisible by $8$ so, if $DEF=320$ then $ABC=18K$ if $K=4$. If $DEF=280$ then $ABC=13K$ if $K=6$

but then I noticed Also $R$ is divisible by $3$ and $9$, so i need the sum of the number to be a multiple of $3$. I have so far

$R=18K320=184320$ where the sum of the digits is $18 $ which is a multiple of $3$ and $9$

I also have:

$R=13K280=136280$ where the sum of the digits is $20$ which is not a multiple of $9$ or $3$. So this number does not work.

So I need help finding two more numbers. Any ideas?

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  • $\begingroup$ Why did you exclude EF = 40 or EF = 60 or EF = 00 (k could be anything) $\endgroup$ – Shailesh Oct 28 '15 at 2:13
  • $\begingroup$ Why is there no possible conditions on $k$.Without it the problem is useless as it can solved easily by trial and a little bit of intuition. $\endgroup$ – tatan Oct 28 '15 at 2:16
  • $\begingroup$ @tatan You can create a condition on $k$ from thinking about the information given. $\endgroup$ – Ian Miller Oct 28 '15 at 2:57
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Since, as you said, 360 divides ABCDEF, 9 divides ABCDEF, so the sum of digits must be a multiple of 9. 0+1+2+3+8=14, so for the sum to be divisible by 9, you must have k=4. From there, you can consider EF=40, and then find multiple of 8 using 1,2,3, and 8. That includes 128 and 328.

EDIT: Apologies, you cannot start with 128 or 328.

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  • $\begingroup$ I believe 312 works. EDIT: Parsing through a few values, you can also end with 20, and with k=4 as before, start with 184. $\endgroup$ – Kevin Long Oct 28 '15 at 2:20
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You excluded some answers with this line:

Also R is divisible by 4 so EF= divisible by 4. but F=0 so, EF=20,80

EF could also be 40 or 60 (or 00 if k is 0).

Worked solution:

Note that 360 is divisible by 9 so R is divisible by 9. The digits of R add up to $14+k$ so $k$ must be 4.

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Look at the last three digits of multiples of 360 and work out which ones can be made using the available digits: 360, 720, 080, 440, 800, 160, 520, 880, 240, 600, 960, 320, 680, 040, 400, 760, 120, 480, 840, 200, 560, 920, 280, 640, 000.

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Look at the five valid end digits combinations and try to create a multiple of 8 from the unused digits.:

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For ###240 can we create a multiple of 8 from 1,3,8. It must end in 8 but neither 138 not 318 is divisible by 8 so it cannot end in 240.

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For ###320 can we create a multiple of 8 from 1,4,8. It must end in 4 or 8. Out of 148, 184, 418, 814 only 184 is divisible by 8 so we have the first solution: 184320.

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For ###120 can we create a multiple of 8 from 3,4,8. Looking at the permutations of 3,4,8 we see that only 384 is divisible by 8. So we have found a second solution: 384120*.

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For ###840 can we create a multiple of 8 from 1,2,3. It must end in 2. 132 does not divide by 8 but 312 does. So we have a third solution: 312840*.

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For ###480 can we create a multiple of 8 from 1,2,3. It must end in 2. 132 does not divide by 8 buy 312 does. So we have a four solution: 312480*.

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For ###280 can we create a multiple of 8 from 1,3,4. It must end in 4. However neither 134 nor 314 are divisible by 8 so there are no more solutions.

The answers are:

184320, 312480, 312840, 384120

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  • $\begingroup$ Possible combinations are 0,1,2,3,8....4,6 are not there....remember k is not being utilised in the last three digits. $\endgroup$ – tatan Oct 28 '15 at 2:14
  • $\begingroup$ Where does it say $k$ can't be in the last three digits? $\endgroup$ – Ian Miller Oct 28 '15 at 2:15
  • $\begingroup$ This is not an answer. should be posted more as a comment. $\endgroup$ – Shailesh Oct 28 '15 at 2:16
  • $\begingroup$ He asked for some help not a solution so this is an answer. $\endgroup$ – Ian Miller Oct 28 '15 at 2:17
  • $\begingroup$ @IanMiller-Why is there no possible conditions on $k$?Without it the problem is useless as it can solved easily by trial and a little bit of intuition. $\endgroup$ – tatan Oct 28 '15 at 2:18

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