Help in algebra calendar

how do I solve an algebra question using the numbers 2,4,6,8 only once each,but it must equal these answers 17,19,25,27,and 31? Please I need help!!

AJ said:

solve an algebra question using the numbers 2,4,6,8 only once each,but it must equal these answers 17,19,25,27,and 31?

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Was this the exact text of the exercise? (It's very poorly written and open to interpretation, is why I ask.)

Thank you.

Eliz.

As always, using the Standard 24 Challenge Rules, we get no solution for 17, 27, and 31.

For 19, try these:

2*(8+(6/4))
(8+(6/4))*2
(6+(4*8))/2
((8*4)+6)/2
2*((6/4)+8)
(6+(8*4))/2
((6/4)+8)*2
((4*8)+6)/2

For 25, try these:

4*(6+(2/8))
4*((2/8)+6)
(6+(2/8))*4
((2/8)+6)*4

If there are other rules contained in the actual problem statement, let's hear them.

No rules, AJ, so everything goes!

2 * 8 + |6/4| = 17

68/4 + 2 = 19

4! + (2+6)/8 = 25

4! + 2 + |8/6| = 27

46/2 + 8 = 31