Denise Margaret Davies
Posted on Sunday, 18 July, 2004 - 01:25 pm:
Using only the digits 1994 I have to find all the numbers 1-100. I must use each digit only once but I must use all four for each solution. I can use any operations but cannot use any that need another digit, for instance squared, without creating a 2 from the existing digits. I have found all the numbers except 87. Can it be done? does anyone know how to do it?
Graeme McRae
Posted on Sunday, 18 July, 2004 - 05:17 pm:
Can you use Floor() and Sqrt()? Can you use place-value, i.e. can you use 91 as a number? If yes, yes, yes, then 91-Floor(Sqrt(Sqrt(4)))-Sqrt(9).
--Graeme
Emma McCaughan
Posted on Sunday, 18 July, 2004 - 09:16 pm:
Can you use decimals? If so, 9/.1 or 9/.1recurring become helpful.
Denise Margaret Davies
Posted on Monday, 19 July, 2004 - 07:31 am:
Yes you can use sqrt and place value but I don't know what floor means so I don't know if it is acceptable. Yes you can use decimal points.
Andrew Fisher
Posted on Monday, 19 July, 2004 - 09:04 am:
How about:

9 x 0.1. = 1

91 - 4 = 87
Denise Margaret Davies
Posted on Monday, 19 July, 2004 - 06:53 pm:
You can use place value but you can't manipulate a number then put it with another. So 9 plus sqrt4 will be 11 not 92.
Andrew Fisher
Posted on Tuesday, 20 July, 2004 - 01:15 pm:
What about:

(91 - 4) x .9.
Alice Thompson
Posted on Tuesday, 20 July, 2004 - 03:15 pm:
or
91-Ö9-   æ
 ú
Ö
  æ
Ö

Ö
Ö{Ö{Ö{Ö{Ö[......4]}}}}
 
 

Andrew Fisher
Posted on Tuesday, 20 July, 2004 - 05:07 pm:
or
(9/0. .
1
 
 )+
Ö
 
9×4
 

Denise Margaret Davies
Posted on Tuesday, 20 July, 2004 - 10:11 pm:
The only one that seems to come up with 87 is 91 - sqrt9 - sqrt4(32 times) from Alice. Thanks Alice.
James
Posted on Tuesday, 20 July, 2004 - 11:18 pm:
There are an infinite number of the square roots in Alice's suggestion. If it was 32, or any other integer, lets say, n, then we would be able to say that.
1 = 4(1/2) n (square rooting a number is the same as raising it to the power 1/2, if we square root it n times then we are raising it to the power (1/2)n )

Then, if we raise each side to the power 2n and we get
12 n = 4

Which is clearly wrong because 1 to the power of any integer power is still 1, so there must be an infinite number of square roots.

James
Andrew Fisher
Posted on Wednesday, 21 July, 2004 - 09:10 am:
My methods do give 87, I promise...

Perhaps you took the small dot to the right of the '9' in my first post and the '1' in my second post to be a typing error. It was in fact my attempt at a 'recurring' symbol.

The first method therefore gives:


(91-4)×0. .
9
 
 =87×1=87

The second method gives:


9/0. .
1
 
 )+
Ö
 
9×4
 
 =(9/(1/9))+   __
Ö36
 
 =81+6 = 87

Alice Thompson
Posted on Wednesday, 21 July, 2004 - 03:54 pm:
If reciprocals are allowed as operations:

9/.1-91/Ö4=90-3=87

Andrew Fisher
Posted on Wednesday, 21 July, 2004 - 03:58 pm:
It's elegant but I fear it might fall foul of:
Quote:

I can use any operations but cannot use any that need another digit

.
Graeme McRae
Posted on Wednesday, 21 July, 2004 - 04:14 pm:
87 = 9/.1-sqrt(sqrt(sqrt(9^4)))