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 From Six To One (Posted on 2006-10-31)
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.

 See The Solution Submitted by Josie Faulkner Rating: 4.4444 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Answer to the question (spoiler) | Comment 5 of 9 |

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) + 82
= 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 2006-11-01 04:29:28

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