Smallest Number with Five 4's

Logic Level 5

Find the smallest positive integer which requires at least five 4's to be written in an expression which is equal to said number.

Allowed operations:

$+, -, \times, \div, \sqrt{ \cdot }$
Brackets
Factorial
Power / Exponentiation
Concatenation of 4's in the expression

Disallowed operators. Everything else, including but not limited to:

Decimal points (since you need a concatenate a 0)
Other mathematical operators like cube roots

An example of how to use these operators is using 4's to create 28 would be $28=4 \times (4+\sqrt{4}+(4 \div 4))$ , which uses five 4's, but you can also do this with two 4's as follows: $28=4!+4$ .

The answer is 39.

1 solution

Sharky Kesa
Feb 27, 2017

I had no good way of determining this easily, apart from just trying each number out until 39. You can skip some of the obvious ones, but here's all of them. If you can find a solution for 39 using four 4's, I'll be very interested with that.

$\begin{aligned} 1&=\frac{4}{4}\\ 2&=\sqrt{4}\\ 3&=4-\frac{4}{4}\\ 4&=4\\ 5&=4+\frac{4}{4}\\ 6&=4+\sqrt{4}\\ 7&=\frac{4!+4}{4}\\ 8&=4+4\\ 9&=\left (\sqrt{4}+\frac{4}{4} \right )^{\sqrt{4}}\\ 10&=4+4+\sqrt{4}\\ 11&=\frac{44}{4}\\ 12&=\frac{4!}{\sqrt{4}}\\ 13&=\frac{4!+\sqrt{4}}{\sqrt{4}}\\ 14&=4\times 4 - \sqrt{4}\\ 15&=4\times 4 - \frac{4}{4}\\ 16&=4\times 4\\ 17&=4 \times 4 + \frac{4}{4}\\ 18&=4\times4 + \sqrt{4}\\ 19&=4!-4-\frac{4}{4}\\ 20&=4!-4\\ 21&=4!-4+\frac{4}{4}\\ 22&=4!-\sqrt{4}\\ 23&=4!-\frac{4}{4}\\ 24&=4!\\ 25&=4!+\frac{4}{4}\\ 26&=4!+\sqrt{4}\\ 27&=4!+4-\frac{4}{4}\\ 28&=4!+4\\ 29&=4!+4+\frac{4}{4}\\ 30&=4!+4+\sqrt{4}\\ 31&=4!+\frac{4!+4}{4}\\ 32&=4\times(4+4)\\ 33&=\frac{\sqrt{\sqrt{\sqrt{4}}}^{4!}+\sqrt{4}}{\sqrt{4}}\\ 34&=4\times (4+4) + \sqrt{4}\\ 35&=4!+\frac{44}{4}\\ 36&=(4+\sqrt{4})^{\sqrt{4}}\\ 37&=4!+\frac{4!+\sqrt{4}}{\sqrt{4}}\\ 38&=4!+4\times 4-\sqrt{4}\\ 39&=4!+4\times 4 - \frac{4}{4} \end{aligned}$

For 39, I don't think there is a better solution. If there is, I'll happily change the answer.

I gave up at 33 :P

Yatin Khanna - 4 years, 3 months ago

Yeah, 33 is a difficult one. You have to note that it is possible to write 64 using two 4's only.

Sharky Kesa - 4 years, 3 months ago

If you allow logs you can go up to infinity: https://www.youtube.com/watch?v=Noo4lN-vSvw

Alex Li - 3 years, 11 months ago

watch that out. I got 1 with five 4's

Rupanshu Shah - 3 years, 10 months ago

You have to use the minimum number of 4's for each integer, as said in the question.

Sharky Kesa - 3 years, 10 months ago

(((root4)^(root4))+4)/(4+4))=1