In the ten problems listed below, your task is to make the number
shown in
bold, using once and once only,
all six numbers, which accompany it.
You may use any mathematical operations you wish, to arrive at the given number.
784 1, 1, 5, 6, 8, 100
327 6, 7, 8, 9, 9, 50
931 3, 4, 7, 8, 10, 75
425 2, 4, 6, 8, 9, 50
489 2, 3, 4, 6, 10, 75
845 4, 7, 8, 9, 9, 25
763 2, 3, 4, 5, 6, 25
599 2, 3, 4, 6, 7, 75
291 4, 8, 9, 9, 10, 100
143 1, 4, 5, 6, 9, 10
Is more than one solution possible for a problem? Please include them if you find some.
There is definitely more than one solution for each. Here are just a few for 784 using 1,1,5,6,8,100:
784 = ((6 * 100) + 5!) + 8^{(1 + 1)}
= (600 + 120) + 8^{2}
= 720 + 64
784 = ((!(5  1)  (6  1)) * 100) + 8!!
= ((!4  5) * 100) + 384
= ((9  5) * 100) + 384
= (4 * 100) + 384
= 400 + 384
784 = (6! + (100/1))  (!5  (8/1))
= (720 + 100)  (44  8)
= 820  36
784 = (8 * 100)  (!(5  1) + (6 + 1))
= 800  (!4 + 7)
= 800  (9 + 7)
= 800  16
784 = (8/1) * (100 + 5  6  1)
= 8 * 98
Here are just a few for 327 using 6,7,8,9,9,50:
327 = (8!! + 9)  ((50 + 6 + 7) + SQRT(9))
= (384 + 9)  (63 + 3)
= 393  66
327 = (50 * 6) + (9 + 8 + 7) + SQRT(9)
= 300 + 24 + 3
327 = !6 + 50 + (9 + 9  8  7)$
= 265 + 50 + 3$
= 315 + 12
327 = (!6 + (50 + 9 + 8  7)) + !(SQRT(9))
= (265 + 60) + !3
= 325 + 2
Edited on November 1, 2006, 3:30 pm

Posted by Dej Mar
on 20061101 04:29:28 