All I see is fours

Logic Level 2

$\large{\begin{aligned} 1 &=& 44 \div 44 \\ 2 &=& 4 \times 4 \div (4 + 4 ) \\ 3 &=& (4 + 4 + 4) \div 4 \\ 4 &=& 4 + (4\times(4-4)) \\ 5 &=& (4 + (4\times4)) \div 4 \end{aligned}}$

Above shows the first 5 positive integers formed by using the four mathematical operators ( $+ \ - \ \times \ \div$ ) only on the digit 4 four times.

What is the smallest positive integer that cannot be represented using these conditions?

Note: You are allowed to join the digits together: $44 + 44$ .

Inspired by a popular recreational puzzle game .

The answer is 11.

2 solutions

Noel Lo
Jul 12, 2015

For me I thought $11 = \frac{44}{4}$ but I need to make use of one more 4 to satisfy the criterion. But if I use another 4, the resultant value will no longer be 11. Now I think of $8 = 4+4$ and I am off by $3$ with two fours to spare. Unfortunately it is impossible to create $3$ with just two fours. Then I try $16 = 4 \times 4$ and this time, I am off by $5$ with two fours to spare. Again it is impossible to create $5$ with just two fours. So $11$ is indeed unworkable. After realising that 1 to 10 are all workable (same method as Ryan Tamburrino ), it is conclusive that $\boxed{11}$ is indeed the smallest number.

Yes indeed , nice proof by showing the impossibility of 11 being made from 2 4's thinking in the terms of the compounds of 2 4s by the use of the operators available.

Now , the question is if there is some way to prove this for all the expressions which use 4 4s with the available operators up to 11 which is indeed harder I think and implies therefore a way to extend the understanding of the particular case relating the impossibility of using 4 4s to make 11 which is more synthetic and harder I think.

A A - 5 years, 1 month ago

Why can't 11 = 4/4 4/4 *"Note: You are allowed to join the digits together" = 1 1
=11

Scott McMahan - 3 years, 8 months ago

There is a way. :) Brute force. Integer results: [-440, -172, -160, -60, -48, -44, -43, -36, -28, -16, -15, -12, -10, -8, -7, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 16, 17, 20, 24, 28, 32, 36, 43, 44, 45, 48, 52, 60, 64, 68, 80, 88, 111, 128, 160, 172, 180, 192, 256, 352, 440, 448, 704, 1776, 1936, 4444] This is obviously not a nice solution. But I have a question, I was thinking on this several times in the past. Where is the limit? I mean, there are problems - even here, on Brilliant - that are usually solved by enumerating all the options. When there are 2 or 3 options, this doesn't really affect the beauty of a solution. Sometimes I see people listing 6-8-10 options in a solution, and somehow we mostly accept it as good enough (in an aesthetic way). Is this always subjective?

Laszlo Kocsis - 3 years, 1 month ago

Ryan Tamburrino
Jun 28, 2015

$6=4+((4+4)\div4)$ $7=4+4-(4\div4)$ $8=4+4-(4-4)=4\times((4+4)\div4)$ $9=4+4+(4\div4)$ $10=(44-4)\div4$ And after about a half hour of thought, I concluded that making $11$ with $4$ fours and standard operations is impossible.

Pi Han Goh - 5 years, 11 months ago

Stealing this. :)

Daniel Liu - 5 years, 11 months ago

All I see is fours

The answer is 11.

2 solutions

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