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# 2, 4, 6, 8 game

How to play: I give a number, you can use the operations $+, -, \times, \div$, powers, roots or factorials, or you can also join numbers to form bigger numbers. Use ALL of the numbers 2, 4, 6, and 8 to make the number I give

Examples: $4 = 8-6+4-2$ $6 = 6+8 \div 4 -2$ $9 = 24 \div 8 + 6$

1. 33
2. 67
3. 25
4. 17
5. 89

The smallest number I couldn't make is 37, somebody help!

Note: using square roots does count the "2" , as if you use $$\sqrt{4}$$ you are using 4 and 2

Note by Takeda Shigenori
2 years, 4 months ago

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The solutions (can I use brackets?): $33=264÷8$ $67=268÷4$ \begin{align} 25&=4!+8÷(6+2) \\&=4!+6÷(8-2) \end{align} $17=4!-(8+6)÷2$ $89=(6!-4×2)÷8$ Helpful tips: Firstly, notice all the numbers given are odd, and 2, 4, 6, 8 are even, the ONLY way to get an odd number here is by division (÷), so the division step must be included in the equation.

Second, the ONLY odd number that uses only two numbers is 3 (6÷2), some other odd numbers must at least use three numbers to achieve.

Third, factorial (!) can never appear in a root because $$n!$$ can never be a perfect root of any number greater than 1.

Fourth, using roots is sometimes number consuming, for example, we can just use $$2$$ instead of $$\sqrt[2]{4}$$, and we can just use $$8$$ instead of $$\sqrt[2]{64}$$.

Thus, there are only 2 forms of solution: the fraction form $\frac{(~~)\pm(~~)}{(~~)}$ or simply addition $|(~~)\pm(~~)÷(~~)|$ (Just fill in the blanks with one or more numbers or nothing, adding operators if needed)

As for 37... it's impossible to get 37 with 2, 4, 6, 8, here's the proof:

Assume it is possible to get 37 with 2, 4, 6 and 8.

Consider the solution is in the addition form. $|\overbrace{(~~)}^{\text{addition part}}\pm \overbrace{(~~)÷(~~)}^{\text{division part}}|=37$

Here, the division part must be an odd number, let's try using only two numbers (6÷2), then $$(~~)=34$$ or $$(~~)=40$$. Now we have to get 34 or 40 using 4 and 8 (since we already used up 2 and 6).

Because $$34=2×17$$, 17 is a prime factor of 34, and there's no way to get a multiple of 17 with 4 and 8 using operators $$×$$, $$÷$$, roots or factorial, so we might try using 4, 8 and operators $$+$$, $$-$$ and factorial to make 34.

After some tries, we found out it's impossible to get 34 with 4 and 8.

As for 40, since 40 is a multiple of 10, if we want to use 4 and 8 with operators $$×$$, $$÷$$, roots and factorial, the product or quotient must have a factor 10. This can only achieved by using $$8!$$, because 8! is way too big than 40, we have to use division, but there are extra factors even when divided by 4 or 4! (for example, $$8!÷4!=8×7×6×5=10×7×6×4$$, there are extra factors 7, 6 and 4), plus there's no way to get 40 with 4 and 8 using $$+$$, $$-$$, factorial, hence, it's impossible to get 40 with 4 and 8.

Now let's try to use 3 numbers in the division part, leaving the last number for the addition part.

Since, the division part must be odd, there must be both 2 and 6 in it, so we can only fill in the addition part with 4, 4!, 8 or 8!.

If we fill in 4 or 4!, then $$(~~)÷(~~)=13,~33,~41,~61$$ , we have to make 13 or 33 or 41 or 61 using 2, 6 and 8, but this is impossible (this is easy to prove).

Similarly if we fill in 8 or 8!, then $$(~~)÷(~~)=29,~45,~40283,~40357$$, there's no way to get 29, 45, 40283, 40357 using 2, 4 and 6.

Consider the solution is in fraction form. $\frac{(~~)\pm(~~)}{(~~)}$ Let's try filling in the denominator with 2, then $$(~~)\pm(~~)=74$$, we have to make 74 with 4, 6 and 8, 74 has a units digit of 4, the ONLY ways to obtain units digit 4 using 4, 6, 8 are $$|\Box8-\Box4|$$, $$\Box8 - 4!$$, $$\Box8+\Box6$$, $$\Box8!+\Box4$$, $$\Box8!+4!$$, $$\Box8!-\Box6$$, $$\Box4×\Box6$$, $$4!×\Box6$$, $$\Box\Box4$$, $$(8-4)×6$$, $$(8-4)!×6$$ (Just fill in the squares with the missing number or nothing), but none of these will equal 74.

Let's try filling in the denominator with 4, then $$(~~)\pm(~~)=148$$, using the same method, the only ways to obtain units digit 8 using 2, 6, 8 are $$\Box2+\Box6$$, $$\Box\Box8$$, $$\Box6!-\Box2$$, $$\Box8!-\Box2$$, but none of these will equal 148.

Using the same method, we could prove that the denominator cannot be 6, 8, 4!, 6! and 8!.

This goes the same for proving the denominator cannot be $$2+4$$, $$2+6$$, $$2+8$$, $$4+6$$, $$4+8$$, $$6+8$$, $$2×4$$, $$2×6$$, $$2×8$$, $$2×4!$$,$$2×6!$$, $$2×8!$$, $$2^4$$, $$\ldots$$

Therefore, 37 cannot be achieved by 2, 4, 6, 8.

Hope this helps. · 2 years, 4 months ago

Yeah you can use brackets, forgot to say that · 2 years, 4 months ago

One more thing: you can use powers · 2 years, 4 months ago