1, 2, 3, 4, 5 = 2017

Can you use the digits 1, 2, 3 4 and 5 (in some order), along with common mathematical operations, to make 2017?

Allowed:
Concatenation of digits Addition, subtraction, multiplication, division Fractions
Exponents, roots
Factorials

What about just the digits 1, 2, 3 and 4? Is that enough to reach 2017?

#Algebra

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Comments

$\frac{(4+3)!}{\frac{5}{2}}+1=\boxed{2017}$

The other case seems quite difficult; given that 2017 is a prime number.

Yatin Khanna - 4 years, 4 months ago

found more by slightly modifying your solution-

$\dfrac{((3!)+2)!}{5 \times 4}+1$ , $\dfrac{(3!)\times(4\times2)!}{5!}+1!$

Anirudh Sreekumar - 4 years, 4 months ago

Very similar to what I did, using 2016+1 :)

Chung Kevin - 4 years, 4 months ago

@Chung Kevin – 2017, being a prime it is hard!.I've gotten to 2016 so many times like- 2016= $2^5(4^3-1)$

Anirudh Sreekumar - 4 years, 4 months ago

@Anirudh Sreekumar – Wow! Can you add all the different ways that you found?

Chung Kevin - 4 years, 4 months ago

@Chung Kevin – hey i got one more for 2017, $(5!+3!)\times 4^{2}+1$

Anirudh Sreekumar - 4 years, 4 months ago

Works fine :)
Any ideas about the 1,2,3,4 case?

Yatin Khanna - 4 years, 4 months ago

@Yatin Khanna – nope still trying :)

Anirudh Sreekumar - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$