# Create number 985 using 1 3 5 7 9

Use $$1, 3, 5, 7, 9$$ in any order to make number $$985$$:

1. You must use all $$5$$ digits $$1, 3, 5, 7, 9$$ exactly once. You can make multi-digit numbers out of the numbers, e.g. $$13$$ or $$975$$.

2. $$+, -, *, /, (), \text{^}, \text{and }!$$ (factorial) are the only allowed functions. Example: factorial may be used more than once, e.g. $$(3!)!=720$$ is acceptable.

• Is the modulo function allowed? (It's used in an answer below) Commented Mar 7, 2023 at 10:45
• Since it never says it's disallowed we must assume it is allowed. I would not advise changing the puzzle after some one came up with a good answer just because it was not the intended one. Commented Mar 7, 2023 at 19:49
• @AndrewSavinykh "Since it never says it's disallowed we must assume it is allowed" I don't think that's valid reasoning - then clearly, the constant-985 function should also be allowed! To me, rule 5 seems like an explicit list of allowed functions. Commented Mar 8, 2023 at 11:34

A small modification to the program from this answer of mine yielded this:

$$985=(3!)!+\frac{7!-5}{19}$$

• Excellent, great finding! Commented Mar 7, 2023 at 19:54

Below equation should give desired result:

$$\mod(5, 3!) * 197 = 5 * 197 = 985$$

• +1 for this solution as well! Commented Mar 7, 2023 at 20:39

Slightly cheeky:

$$-5-\dfrac{(-9)!}{(1-3!-7)!}$$

"Justification"

$$\displaystyle\lim_{x\to-n} \frac{\Gamma(x+1)}{\Gamma(x)} = -n$$

The same method allows for a variation of @Aman's answer

$$-197 \times (-5)! / (-(3!))!$$

A cheeky, but ingenious way:

We do $$7+1$$ to make $$8$$ and then use $$9$$, $$8$$ (The number from the previous equation) and $$5$$ to make $$985$$.
Here's the cheeky part: We use the last digit ($$3$$) by doing $$x = 3$$. Then we can do $$985 + (x - x)$$ to get $$985$$.
$$x = 3$$ isn't a function!

Full equation ($$answer$$ means answer):

$$answer = (985) + (x - x), x = 3, 1 + 7 = 8$$