Anyone want to solve a math puzzle?

Question

1, 2, 3, 4, 5,6

a. Rearrange the numbers so that when adjacent, they differ by at least three and the leftmost # is as small as possible.

b.arrange the numbers so that the adjacent numbers is less than or equal to seven and the leftmost # is as small as possible.

c. arrange the numbers so that adjacent numbers form two digit numberrs that cannot be divided evenly by 3, 4, and 5.

Answer 1

a. 3 has to be next to 6 so 3 6 2 5 1 4

b. 6 has to be next to 1 and either at the end or beginning so 3 4 2 5 1 6

c.  the 5 would have to come first since otherwise there would be a 2 digit number ending in 5, and anything but 3 next would not work, so

5 3 1 4 6 2

Answer 2

a)

The only number differing by at least 3 from 3 is 6.

So 36 must be on left (or reversed on right -

we might reverse the whole thing at the end).

Next can be 1 or 2.

If it's 1, then next can be 4 or 5, but no next if it's 4, so 5

3615 then only 2 4 .. no good

Back up to 362 then 514

Since 3 < 4, we keep 3 on the left: 362514

b)

6 can only partner with 1, so it must be on the end,

and since all others are smaller, it must be on the right.

Ending is 16.

5 can only partner with 1 and 2, so it must be next

then 2

2516

then 4 and 3 in either order, choose 34 to have smallest left end

342516

c)

5 must go on the left (otherwise some 2 digit number ends in 5)

next cannot be 6 or 1 or 4 or 2, it must be 3

53

If it's 4, following 534:

5342 does not work

5346 neither 612 nor 621 works.

5341 cannot be followed by 2 or 6.

If 1 follows 3, following cannot be 2 or 6, must be 4

5314 then 62 that works, 26 does not.

531462

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