
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 1st, 2018, 03:44 AM  #1 
Newbie Joined: Feb 2018 From: Ocala Posts: 2 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: 11,647 Thanks: 740 
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: 229 Thanks: 81 
$$(1.2+1.3+0+0+0)^{1.4+1.5} = 14.25...$$ Biggest one I found: $$(1.3  1.2)^{0+01.41.5} = 794.32$$ 
February 1st, 2018, 06:14 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,647 Thanks: 740 
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: 229 Thanks: 81 
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(n2)(n4) ...). So can you get something arbitrary large without the factorial? 
February 1st, 2018, 07:16 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,557 Thanks: 1480 
$\displaystyle \frac{1.2 + 0 + 0 + 0}{1.5(1.4 1.3)} = 8$
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: 11,647 Thanks: 740 
I knew Skip would get it!

February 1st, 2018, 12:53 PM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,647 Thanks: 740  Quote:
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 
Newbie Joined: Feb 2013 From: London, England, UK Posts: 28 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. 

Tags 
math, problem, solve, unsolvable 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
10 years old math problem  please help me solve it  mihaela  Elementary Math  3  November 24th, 2017 03:17 PM 
How can I solve this math problem?  DerGeraet  Linear Algebra  2  November 10th, 2011 09:07 AM 
an unsolvable problem  UltraMath  Algebra  3  September 4th, 2011 10:54 AM 
Trinagle Problem. Math Award Winner Can't Solve, IMPOSSIBLE?  nqavisid  Algebra  4  November 14th, 2008 01:28 PM 