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-   -   Unsolvable math problem ... can you solve it? (http://mymathforum.com/number-theory/343324-unsolvable-math-problem-can-you-solve.html)

Jumbo February 1st, 2018 03:44 AM

Unsolvable math problem ... can you solve it?
 
Hi, I'm new here and was wondering if anyone could solve this math challenge. On the surface this question seems basic indeed. Most would instantly say "there's no solution possible", but I'm not so sure. Here it is ......

I will list 7 numbers (including zeros). My question is .... is there any way possible for these numbers to equal 8 or more when added or multiplied or using any other math technique? Each number can only be used "once", but can be used in any order. The numbers also can't be changed (for example ... you can't place 2 zeros after the 1 in 1.2 and then call it 100.2).

The 7 numbers are:

1.2
0
1.3
0
1.4
0
1.5

If they are added they come to 5.4 and multiplied they come to 3.276. Both of these numbers are below 8.

Can any of you, using creative math, make these numbers total 8 or more?

Denis February 1st, 2018 05:28 AM

Since any other math technique is allowed:
(1.2 + 1.3 + 1.5)! = 4! = 24
24 + 1.4 + 0 + 0 + 0 = 25.4

Micrm@ss February 1st, 2018 06:05 AM

$$(1.2+1.3+0+0+0)^{1.4+1.5} = 14.25...$$

Biggest one I found:

$$(1.3 - 1.2)^{0+0-1.4-1.5} = 794.32$$

Denis February 1st, 2018 06:14 AM

Since 0! = 1:

(1.3 - 1.2)^(-0! - 0! - 0! - 1.4 - 1.5) = 794328.234...

Micrm@ss February 1st, 2018 07:06 AM

Fine, but then we can just adjust your post

0+0+0+1.4+(1.2 + 1.3 + 1.5)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
to get arbitrary large numbers (note that I do not use the convention here n!! = n(n-2)(n-4) ...).

So can you get something arbitrary large without the factorial?

skipjack February 1st, 2018 07:16 AM

$\displaystyle \frac{1.2 + 0 + 0 + 0}{1.5(1.4- 1.3)} = 8$

Denis February 1st, 2018 10:22 AM

I knew Skip would get it!

Denis February 1st, 2018 12:53 PM

Quote:

Originally Posted by Micrm@ss (Post 587943)
Fine, but then we can just adjust your post

0+0+0+1.4+(1.2 + 1.3 + 1.5)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
to get arbitrary large numbers (note that I do not use the convention here n!! = n(n-2)(n-4) ...).

So can you get something arbitrary large without the factorial?

RIGHT!
I wonder why the problem is not simply stated as:
arrange 1.2, 1.3, 1.4 and 1.5 using only *, /, +, -
to get result of 8.

Jopus February 2nd, 2018 03:25 PM

Jumbo Response
 
Hi Jumbo,

I like your question. I like skipjack's response. Very exact, but only equal to 8, not greater than it. I would offer by way of contrast the following:

First, rewrite as follows:

1.4 = 14/10;
1.5 = 15/10;
1.3 = 13/10;
1.2 = 12/10;

Now:

(14/10 + 15/10) = (29/10)

&: (13/10)^0 = 1
(12/10)^0 = 1

Adding: 1 + 1 = 2.

Squaring: (29/10)^2 = 841/100 = 8.41 ( 0.41 greater than 8 )

Take care, and all the best skipjack.

(Incidentally, I have a spare zero also left over, which is very good)

Maybe you should follow the path of the wise and noble buddha that is Mr CRGreathouse and voluntarily retire yourself from all forums from now on. The times, they are a changing, and there's a new song 'a blowin' in the wind, as Bob Dylan famously didn't say.

Adios Hoez.

Collag3n February 27th, 2018 12:15 PM

Quote:

Originally Posted by Jopus (Post 588005)
... Very exact, but only equal to 8, not greater than it....

did you try to find the smaller value greater than 8? like $(1.3+1.2^0)^{(1.5+1.4^0)}$?
Exponenet authorized? $\frac{1.3^0+1.4}{1.5-1.2}$


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