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 October 17th, 2018, 03:17 AM #11 Member   Joined: Sep 2017 From: Saudi Arabia Posts: 37 Thanks: 1 Using the digits; 1,2,3,4,5,6,7, any number to be divisible by 4 must end with 12,16,24,32,36,52,56,64,72, or 76. In each of these 10 cases, we have 5 unused digits, which can be permuted in 5! ways (or 120 ways). Since every case has 120 ways, and we have 10 cases, therefore the total number of ways = 120 x 10 = 1200 ways.
October 17th, 2018, 04:14 AM   #12
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Joined: Jul 2018
From: Georgia

Posts: 28
Thanks: 7

Quote:
 Originally Posted by Denis Answer will be: none!!
Not sure if you're joking or if I'm missing something.

We can, of course, form multiple numbers evenly divisible by six using the 7 integers supplied. They must all be even numbers, but once we get to numbers of 100 and above then we end up with 3 different sets.

Ignoring, for the moment, the restriction regarding not duplicating integers, except for the the last two digits (but remembering that we don't have a zero), the possibilities for numbers of the form 1xx (or 4xx or 7xx) would be:

114 (by our rules we can't do that with either 100 or 400, but it does work for 700)
126
132
156
162
174

For 2xx (or 5xx), the possibilities are:

216
234
246
252
264
276

With 3xx (or 6xx), it's:

312
324
336
342
354
372

But as we get to 4, 5, 6 or 7 digit numbers it gets more complicated. For example, the non-duplication rule regarding the last two digits for the second set above eliminates four 2-digit possibilities for any larger number containing a 6, but only one 2-digit possibility for a larger number containing a 2. But the 3rd set eliminates 4 possibilities for 2 and only one possibility for 6.

And at this point I'm thoroughly confused. Is there any straightforward way to calculate the total number of valid numbers for this problem?

October 17th, 2018, 04:28 AM   #13
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Joined: Oct 2013
From: New York, USA

Posts: 639
Thanks: 85

Quote:
 Originally Posted by Denis Answer will be: none!!
The sum of the numbers from 1 to 7 is 28. To be a multiple of 6, the sum of the digits must be a multiple of 3. Therefore a seven-digit number with the numbers 1 to 7 once each cannot be a multiple of 6. If you don't have to use every number, you can make multiples of 6 such as 123,456. If you are talking about only seven-digit numbers and RichardJ is including numbers with fewer digits, that explains the disagreement.

October 17th, 2018, 04:40 AM   #14
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Joined: Jul 2018
From: Georgia

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Quote:
 Originally Posted by EvanJ The sum of the numbers from 1 to 7 is 28. To be a multiple of 6, the sum of the digits must be a multiple of 3. Therefore a seven-digit number with the numbers 1 to 7 once each cannot be a multiple of 6. If you don't have to use every number, you can make multiples of 6 such as 123,456. If you are talking about only seven-digit numbers and RichardJ is including numbers with fewer digits, that explains the disagreement.
Ahh, thank you. I didn't know (or more likely had forgotten) that fact about the sum of the digits.

That probably explains my confusion (well, some of it, anyway).

 October 17th, 2018, 04:55 AM #15 Newbie   Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7 Actually, just ignore everything I said. At this point I'm sure it's a pile of nonsense and not at all relevant. Sorry.
 October 17th, 2018, 05:34 AM #16 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 Ok ok Richard...don't commit suicide... Thanks from RichardJ

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