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Using 2,3,4,5,6,7 and basic mathematical operations form the number 1088.

Every digit must be used exactly once (neither zero nor greater than one times). Concatenation of digits (such as 53) is not allowed. Each operator can be used multiple times or not at all in any combination. This removes duplicates and creates it as having a unique solution.

Note:

Allowed Operations: $+,-,/,*, {}^{\displaystyle\large\hat{}} ,!,\sqrt{}$

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    $\begingroup$ Must we use all provided numbers or can we use only a subset of them? Do you allow repeats? Can we combine numbers to form multi-digit numbers (e.g. 23, 57 etc)? Also, is ^ exponentiation or XOR? $\endgroup$
    – user12205
    Commented May 11, 2015 at 12:41
  • $\begingroup$ @ace Every digit exactly once. $\endgroup$
    – RE60K
    Commented May 11, 2015 at 13:05
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    $\begingroup$ ADG, you should explicitly state (edit it in) that parentheses are allowed if they are allowed. $\endgroup$
    – Olive Stemforn
    Commented Jun 20, 2015 at 16:56

5 Answers 5

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Update: This 4-operator answer is now disallowed by OP after rule change. The 6-operator one still works though.

An answer that uses 4 operators only (assuming concatenation to form multi-digit numbers is allowed):

$$ 523 \times \sqrt4 + 6 \times 7 = 1088 $$

Assuming we can use parentheses and they don't count as operators, then we have 6 operators with

$$ 7^3 + 6! + 5 ^ {(4-2)} = 1088 $$

If we can't use parentheses, then

$$ 7\;3\;{}^\wedge\;6\;!\;+\;5\;4\;2\;-\;{}^\wedge\;+ $$

which is the same thing in reverse Polish notation.

The above assumes that the ^ operator is for exponentiation, not XOR.

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  • $\begingroup$ You can remove the parentheses in any case since $5^{4-2}$ is unambiguous. $\endgroup$
    – Lawrence
    Commented May 12, 2015 at 15:06
  • $\begingroup$ @Lawrence I know, but the OP mentioned the ^ operator, not exponentiation. If we are to use the ^ operator, we would need to write 5^(4-2) with parentheses needed. I'm not sure how the OP wants his answers. $\endgroup$
    – user12205
    Commented May 12, 2015 at 15:13
  • $\begingroup$ I see - good call then :) . $\endgroup$
    – Lawrence
    Commented May 12, 2015 at 15:14
  • $\begingroup$ From OP: "No concatenation" Your solutions using 523 and 73/542 are invalid. $\endgroup$
    – Engineer Toast
    Commented May 12, 2015 at 15:38
  • $\begingroup$ @EngineerToast The OP stated that rule after my submission. (I asked him in the comments but he did not reply.) Also, the last entry is the same as the second but written in reverse Polish notation. $\endgroup$
    – user12205
    Commented May 12, 2015 at 15:43
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The Answer is

7!/5 + 6!/(2+3+4)

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Here is another possible solution:

$(5!/4 \ + \ 3)^2 \ - \ 7 \ + \ 6 \ = \ 1088$

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One other possible answer: $$6! \times \frac{3}{2} + 5 + 7 - 4 = 1088$$

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Simple answer:

$1088=2^{(2*5)}+64$

If we must use every digit once....

$1088=2^{(3+7)}+64$

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  • $\begingroup$ Every digit exactly once. $\endgroup$
    – RE60K
    Commented May 12, 2015 at 14:58
  • $\begingroup$ @ADG Minor edit; solved it! $\endgroup$
    – ghosts_in_the_code
    Commented May 12, 2015 at 15:20
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    $\begingroup$ sorry that I am not able to tell you the problem exactly. I use every digit once and exactly once, you have no 5. No concatenation like 64. But you can use an operator many times. $\endgroup$
    – RE60K
    Commented May 12, 2015 at 15:27

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