puzzle

[ijabbott]

(06-29-2018 04:27 AM)Thomas Klemm Wrote:  

(06-29-2018 03:40 AM)Don Shepherd Wrote:  How many 4-digit arrangements would you have to check?

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12 = 4! / 2!

1223
1232
1322
2123
2132
2213
2231
2312
2321
3122
3212
3221

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6 if you initially discount the even numbers by inspection - there are only two choices for the units digit.

(You have filtered out the even numbers out in the next step.)

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Quote:Is there a way to check other than an electronic prime factor finder or our old friend, brute force?

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Even numbers are rarely prime:
1232
1322
2132
2312
3122
3212

Check alternating sum of digits:
2123 -> 2 - 1 + 2 - 3 = 0 => divisible by 11: 2123 = 11 * 193
2321 -> 2 - 3 + 2 + 1 = 0 => divisible by 11: 2321 = 11 * 211

Brute force:
2231 = 23 * 97

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You could also do an initial sum over the digits:

1+2+3+2 = 8

Not a multiple of 3, so it is worth considering other checks for primality. If the given number was 1233, it would fail this initial test - all arrangements would be divisible by 3.

Quote:That was fun.

I really wish there was a [spoiler]spoiler[/spoiler] tag!