My Math Forum Unsolvable math problem ... can you solve it?

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 February 1st, 2018, 03:44 AM #1 Newbie   Joined: Feb 2018 From: Ocala Posts: 7 Thanks: 0 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? Last edited by Jumbo; February 1st, 2018 at 04:14 AM.
 February 1st, 2018, 05:28 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,789 Thanks: 970 Since any other math technique is allowed: (1.2 + 1.3 + 1.5)! = 4! = 24 24 + 1.4 + 0 + 0 + 0 = 25.4
 February 1st, 2018, 06:05 AM #3 Senior Member   Joined: Oct 2009 Posts: 687 Thanks: 223 $$(1.2+1.3+0+0+0)^{1.4+1.5} = 14.25...$$ Biggest one I found: $$(1.3 - 1.2)^{0+0-1.4-1.5} = 794.32$$
 February 1st, 2018, 06:14 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,789 Thanks: 970 Since 0! = 1: (1.3 - 1.2)^(-0! - 0! - 0! - 1.4 - 1.5) = 794328.234...
 February 1st, 2018, 07:06 AM #5 Senior Member   Joined: Oct 2009 Posts: 687 Thanks: 223 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?
 February 1st, 2018, 07:16 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,096 Thanks: 1905 $\displaystyle \frac{1.2 + 0 + 0 + 0}{1.5(1.4- 1.3)} = 8$ Thanks from Sebastian Garth Last edited by skipjack; February 1st, 2018 at 07:20 AM.
 February 1st, 2018, 10:22 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,789 Thanks: 970 I knew Skip would get it!
February 1st, 2018, 12:53 PM   #8
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Posts: 13,789
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Quote:
 Originally Posted by Micrm@ss 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.

 February 2nd, 2018, 03:25 PM #9 Member   Joined: Feb 2013 From: London, England, UK Posts: 37 Thanks: 1 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.
February 27th, 2018, 12:15 PM   #10
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Joined: Sep 2017
From: Belgium

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Quote:
 Originally Posted by Jopus ... 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? $\frac{1.3^0+1.4}{1.5-1.2}$

Last edited by Collag3n; February 27th, 2018 at 12:33 PM.

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