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Allowed Operations:

Addition, Subtraction, Multiplication, Division, Exponentiation. Right and Left Bracket use permitted.

Express $2222$ in $2$ different ways using all the digits 1 to 9 only once in each expression.

$2222$ =

$2222$ =

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  • $\begingroup$ So you can't use zero? $\endgroup$
    – Duck
    Commented May 31, 2019 at 2:23
  • $\begingroup$ No..only 1 to 9,,all need to be used without repeating $\endgroup$
    – Uvc
    Commented May 31, 2019 at 2:28
  • $\begingroup$ And also, can you make, for example, 4 and 5, 45? $\endgroup$
    – Duck
    Commented May 31, 2019 at 2:29
  • $\begingroup$ If you mean it as forty five..sure $\endgroup$
    – Uvc
    Commented May 31, 2019 at 2:31
  • $\begingroup$ For example: one of the terms could be..(5/1) ^ 3 $\endgroup$
    – Uvc
    Commented May 31, 2019 at 2:36

3 Answers 3

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I'm going to try:

$2222=1987+234-5+6$

and

$2222=1986+234-5+7$

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Just a little bit after @JonMark Perry

$(49+52) * (8+3) * (7-6+1)$

and technically different

$(42+59) * (3+8) * (7-6+1)$

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  • $\begingroup$ Good one......... $\endgroup$
    – Uvc
    Commented May 31, 2019 at 18:48
  • $\begingroup$ Just some prime factorization... $\endgroup$
    – Duck
    Commented May 31, 2019 at 22:20
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Solution 1:

To get a “big” number like $2222$, I decided to use powers.

$2^{11}=2048$ is close to $2222$ but is a bit too small.
I need to increase my solution by $2222-2048=174$.
I can express the power of $11$ as $3+8$ or $4+7$ or $5+6$.

Examining the $3+8$ case, the unused digits are: $1,4,5,6,7,9$.
We can easily make the number $174$ using the digits $1,7,4$.

To deal with the remaining digits $5,6,9$ we use two facts:
$1^n=1$ and multiplying by $1$ doesn’t change the value of an expression.

Finally, we get:

$2222=2^{3+8}+174 \times (6-5)^9$

Solution 2:

Considering powers of $3$, I got a nicer solution:

$2222=3^7+1+2+4+5+6+8+9$

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