Solving Equality w/ 2,3,4,5,+,=: Two Ways?

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The discussion centers on creating an equation using only the numbers 2, 3, 4, and 5, along with the operators + and =, with the stipulation that each number and operator must be used exactly once. Initial suggestions included various combinations, but the constraints of using each number and operator only once led to confusion. Participants explored the possibility of interpreting adjacent numbers as multiplication, but this was deemed outside the original parameters. The consensus emerged that finding multiple valid equations under these strict conditions is challenging, with one proposed solution being 4 + 5 = 3². However, this solution uses an exponent, which violates the requirement of using only the specified operators. Ultimately, the group concluded that it may be impossible to create a valid equation that meets all the criteria without exceeding the allowed symbols and operations.
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how can you write an equality ONLY using 2, 3, 4, 5, +, and =? note ALL must be used. I have done this problem and it is easy but i was wondering if there are two ways to do it. Any ideas?
 
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2 + 5 = 3 + 4

There are infinitely many ways of doing it, providing you don't limit the number of symbols used. I believe this, and rearrangements of this, is the simplest.

2 + 3 + 4 = 5 + 4

2 + 2 + 2 + 3 = 4 + 5

2 + 5 = 3 + 4 = 2 + 2 + 3

...

ad infinitum
 
i believe i misstated the question. the numbers 2 3 4 and 5 can be used only once as with the = and + sign. sorry about that so

3 + 4 = 2 + 5 would NOT work because you would be using the plus sign twice
 
Hmmm, would having two numbers next to each other count as multiplying? For instance 3x is seen as multiplying 3 times the value of x, so could the answer be written 5 + 3 = 2 4 with the statement that 2 and 4 are being multiplied?
 
first attempt: 4+5=3²
 
thats what i got too :) but do yuo know any other ways?
 
i don't think that there are any more...
look:
we have '+' and '=' operators, thus 3 free slots that are to be filled with at least one digit each. so 3 of 4 given didits are bound. that is we have only one free digit left. now, there are not that much possibilities in choosing the numbers for each slot AND choosing the position for the "free" digit. So we can just check all the possibilites for consistency...
 
hemmul said:
first attempt: 4+5=3²

can this equation not be written more appropriately as

4+5=3^2

thus we are using more than 2 operators. in fact in this case it is not possible to get an equation as with 2 operators we can use only 3 numbers...


thus the only way of using 4 digits would be if we use a 2 digit no. like 24, or 35, or something like that,, but then it is not possible to get any equation. :smile:
 
im assuming that writing 2^3 on paper by hand would just be the numbers 2 and 3
if i am not counting operators, just what's written :)

so anyone know another way? i believe its impossible but i am not sure...
 
  • #10
T@P said:
im assuming that writing 2^3 on paper by hand would just be the numbers 2 and 3
if i am not counting operators, just what's written :)
In that case 2 4 should be ok also.
 
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