hi ok the details are you cano nly use the numbers 1,4,8,and 9 and you can only use the once in each secunse. You have to use all the numbers, you can't use a 2 diget number. You can use any mathimaticl epression.
the numbers I need are 1, 7, 16, 26, 30, 34, 38, 39, 47-49, 51, 57, 61-66, 71-74, 78-83, 87-89, 93-95
If you can get any of thouse number it would really help and I would like it for mon sept 24 if you can
THANK-YOU for looking/answering my question
2007-09-20 11:51:46 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I got 16 2 different ways:
16 = 1 + 4 + 8 + √9
and
16 = (9-8+1)^4..
Now recall that 3! = 3*2*1 = 6 and 4! = 4*3*2*1 = 24
so
4! + 9 - 8 + 1 = 24 + 2 = 26.
Here are some more for you:
(4! + √(8+1) + √9 = 30
4! + √9 +8 - 1 = 34
(4!+√9! +8)*1 = 38
8(9-4) -1 = 39
8(√(9*4) - 1 = 47
8*(√(9*4)*1 = 48
8(4+1) + 9 = 49
Update: Here are the rest of them for you:
4! + (8+1) * √9 = 51
8(9 - √4) + 1 = 57
9(8-1) - √4 = 61
8(9-1) - √4 = 62
(8-4)^√9 - 1 = 63
((8-4)^√9)* 1 = 64
9(8-1) + √4 = 65
8(9-1)+ √4 = 66
8*9 + 1 - √4 = 71
(8-4)! * √9^1 = 72
8*9 + √4 - 1 =73
(8*9 + √4)*1 = 74
8(9+1) - √4 = 78
9(8+1) - √4 = 79
√9^(8-4) - 1 = 80
(√9^(8-4))* 1 = 81
√9^(8-4) + 1 = 82
9(8+1) + √4 = 83
8(9+ √4)-1 = 87
√9^4 +8 - 1 = 88
(9 ^ √4+ 8)* 1 = 89
(4! + 8-1)*√9 = 93
(8*√9! - 1)* √4 = 94
8*9 -1 + 4! = 95.
Finally got 38, 93 and 94
All done!
Good luck with your project!!
2007-09-20 14:15:13 · answer #1 · answered by steiner1745 7 · 0⤊ 0⤋
What the heck is a secunse? Anyway...
1 = (9 - 8) * ((√4) - 1)
7 = (9 - (8/4))*1
16 = (9-1)*(8/4)
2007-09-20 12:06:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋