expression for 29 using 1, 2, 3, 4, *, /, +, -, powers

[Blessedn94]

:? I'm not really sure where I should ask this, it is for my daughter. She needs to find the sum of 29 using 1,2,3,4. she can not repeat the numbers, however they can be multiplied, divided, added, subtracted and you can use exponents. Thank you.

 

[jwpaine]

Surely you can help her by trial and error.

Example: does 1+2+3+4 = 29? Does 1 * 2 * 3 * 4 = 29? Try different things until both sides are equal.

Best of luck!
John

 

[Loren]

You might figure out how to get 28 using 2, 3 and 4, then add 1. Or figure out how to get 27 using 1, 3 and 4 then add the 2. Or maybe get 30 using 2, 3 and 4 and subtract 1, etc.

 

[TchrWill]

Re: sum of 29

:? I'm not really sure where I should ask this, it is for my daughter. She needs to find the sum of 29 using 1,2,3,4. she can not repeat the numbers, however they can be multiplied, divided, added, subtracted and you can use exponents. Thank you.

This type of problem is strictly a trial and error process.

How about 32 - (4 - 1) = 29

How about 31 - (4 - 2) = 29

 

[skeeter]

\(\displaystyle 2^{4+1} - 3\)

 

[Denis]

Keeping 'em in order:
(1 + 2)^3 + sqrt(4) : since sqrt(4) = 4^1/2

 

[Sdalgliesh]

Simple

Keeping 'em in order:
(1 + 2)^3 + sqrt(4) : since sqrt(4) = 4^1/2

It's quite simple, think outside the box... using 1,2,3,4 make 29..

42 -13 =29

This uses all numbers...

 

[mmm4444bot]

… 42 -13 =29

This uses all numbers...

Hi Sdalgliesh. You have applied concatenation, but the instructions don't list concatenation as an option. The numbers 42 and 13 are not an option, as I read it, because {1,2,3,4} are described as numbers, not digits.

Without repeating answers already posted, can you find a way to get 29 with only addition, subtraction, multiplication, division and/or exponentiation, using all four numbers 1, 2, 3 and 4?

 

[lookagain]

Keeping 'em in order:
(1 + 2)^3 + sqrt(4) : since sqrt(4) = 4^1/2

It's quite simple, think outside the box... using 1,2,3,4 make 29..

42 -13 =29

This uses all numbers...

Sdalgliesh,

also, there are a couple notes about the post by Denis.

1) The square root is not allowed for this problem.

2) "4^1/2" is an incorrect form, because it does not follow the Order of Operations.

It should be "4^(1/2)" or something equivalent with other grouping symbols.

- - - - - - - - - - - - - - - -

I see mmm4444bot edited TchrWill's post back in 2007, but mmm4444bot's comment referring to
concatenation could have been made back right after TchrWill's post, as well as it was made just
recently to Sdalgliesh.

 

[Denis]

I'll go stand in the corner for 4^(1/2) minutes...

 

[mmm4444bot]

I see mmm4444bot edited TchrWill's post back in 2007 …

No. TchrWill posted it back in 2007; I edited just recently, when I noticed a broken quote.

… mmm4444bot's comment referring to concatenation could have been made back right after TchrWill's post …

I was out of town. :cool:

 

[mmm4444bot]

I'll go stand in the corner for 4^(1/2) minutes...

I don't understand why lookagain suggests rational exponents are not allowed. I'd argue for your release, buddy, but ya did repeat numbers 1 and 2, and Blessedn94 said no repeats.

 

[lookagain]

I don't understand why lookagain suggests rational exponents are not allowed.

No, I am not suggesting that.

"4^(1/2)" combined with a "3" somewhere, if they would equal 29, hypothetically speaking, are okay.

"\(\displaystyle \sqrt{4}\ \)" is not okay, because using square roots was not mentioned as being allowed in the rules.

 

[mmm4444bot]

… 4^(1/2) … [is] okay … \(\displaystyle \sqrt{4}\) … is not okay …

Thanks for the clarification.

 

[Steven G]

I'll go stand in the corner for 4^(1/2) minutes...

I think you should go in the corner for 4^1/2 minutes instead.

 

[Denis]

I think you should go in the corner for 4^1/2 minutes instead.

Thanks buddy...that's only 2 minutes:cool: