Help in algebra calendar

[AJ]

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!!

 

[stapel]

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?

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

Thank you.

Eliz.

 

[tkhunny]

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.

 

[Denis]

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