Making 35, 37, 39, and 41 with four 4s

[Joseph]

I need to make 35,37,39,and 41 using four 4's.

 

[cookiemonster]

If the only operations you can perform on the fours is addition, subtraction, multiplication, and division, then you can't get any of them.

You'll always get an even number, regardless of the operations you use or the order you use them in.

cookiemonster

 

[Bob3141592]

If the only operations you can perform on the fours is addition, subtraction, multiplication, and division, then you can't get any of them.

You'll always get an even number, regardless of the operations you use or the order you use them in.

cookiemonster

No, you can use two of the fours to produce one, as in 4 + 4 + 4/4 = 9. Or 4*4 + 4/4 = 17. Subtracting the quotient gives 7 and 15 respectively.

Offhand, given the four function constraint, I don't see a way to get a higher odd number than 17.

Can we use the four fours as strings instead of individual digits? 444/4 = 111, an odd number. 44/4 + 4 = 15. (44/4)^4 = 14,641, a very big odd number.

On the other hand, can we use factorials? 4! is 24, so 4! + 4! - 4/4 = 47. Drats, it's not on the target list. (4 * 4! - 4) / 4 is 23. Still no go.

Sorry, but I don't see a trick to get the desired numbers.

 

[arildno]

This is to cheat, but: ((4*4)4)/4=164/4=41

 

[JasonRox]

Aren't all numbers primes or products of primes?

 

[JasonRox]

I came pretty close for one of them.

$$(\frac{4^2}{ \sqrt{1/4}}) - 4^0 = 31$$

Can you see what I did?

Here's another one.

$$(\frac{4}{ \sqrt{1/4}}) 4 - 4^0 = 31$$

$$(\frac{4^2}{ \sqrt{1/4}}) + 4^0 = 33$$

 

[JasonRox]

Another one.

$$44 - 4 - 4^0 = 39$$

 

[JasonRox]

I got to go, so I'll do the rest when I get back.

Use my tricks and you are good to go.

 

[JasonRox]

$$44 + 4^0 - 4 = 41$$

Got back from lunch. Time to study boring accounting.

 

[cookiemonster]

Aha! I forgot to consider 4/4 = 1.

Good point.

cookiemonster

 

[Bob3141592]

I came pretty close for one of them.

$$(\frac{4^2}{ \sqrt{1/4}}) - 4^0 = 35$$

Can you see what I did?

Here's another one.

$$(\frac{4}{ \sqrt{1/4}}) 4 - 4^0 = 31$$

$$(\frac{4^2}{ \sqrt{1/4}}) + 4^0 = 37$$

I'd presumed that using other numerals like 2 and 0 for exponents wouldn't be allowed. At least that's the case in problems of this sort I've seen before. Non-numeric mathematic symbols like square root are OK, but a cube root isn't be, since you have to use a numeral to specify it. If you can use a 4^2 to mean squared, why can't you say 4*5 or anything else?

 

[JasonRox]

I put squared to make it easier. If you count again, you will see there is four.

Remember all numbers have exponent 1's.

If I'm allowed square roots, I did nothing wrong. Square root is the same as exponent 1/2.

I didn't break any rules if you allow 1/2.

I believe that you are restricted to the number one. If you add 1/2 and 1/2 you get 1, and since you allowed 1/2, I can manipulate it to do other things. Here is how it works:

-1/2-1/2=-1 (I used 1/4 instead of 4^-1, because it didn't work for some reason)
1/2-1/2=0
1/2+1/2=1 (The regular exponent.)

See I broke no rules. :)

 

[JasonRox]

More...

$$4! + 4^2 + 4^0 = 41$$

There is 4 Four's.

 

[JasonRox]

Let's here the opposite of Fermat's Last Theorem!

$$x^n + y^n = z^n$$, is possible in infinite amounts for n<2.

 

[Bob3141592]

Was this a class assignment? If so, let me know if changing a four to a one by raising it to the zeroth power is acceptable. I'd be surprised, but I'd like to know.

Also, earlier you wrote:

Aren't all numbers primes or products of primes?

All natural numbers are, that is, positive integers. Obviously negative numbers and fractional numbers aren't.

Further, each natural number is either prime or the product of primes in a unique way. There is one and only one such representation for each. That's the fundamental theorem of arithmetic.