Making 1-50 from 2016

Question

Make the numbers 1-50 using the numbers 2 0 1 6 in the given order.

1.You must use all four digits.

2.You may not use any other numbers.

3.You may use +, -, x, ÷, square root, squaring and cubing, exponentiation, parentheses, brackets, or other grouping symbols.

Can you form two-digit numbers, e.g. 20+16=36? Please edit your question accordingly. — Matsmath, Sep 11 '16 at 11:39
In an earlier revision of the question, you said raising to the powers of 2 and 3 was allowed. Is this still allowed, even though 3 isn't one of the four digits provided? — Rand al'Thor, Sep 11 '16 at 12:03
Are we allowed to use module (%), or (|), and (&), xor (~), byte-shifts (>> and <<), etc., since you state all operations are allowed? — Kevin Cruijssen, Sep 11 '16 at 14:22

score 12 · Answer 1 · edited Apr 13 '17 at 12:50

12

Here's the complete list from 1 to 50. Some of them can be done in a 'pure' way, with no numbers appearing in the expression except the four specified. Others require squaring or cubing (which the OP said is permitted), and I've marked these as such.

$2*0+1^6$
$2+0*16$
$2+0+1^6$
$20-16$
$2*0-1+6$
$2*0*1+6$
$2*0+1+6$
$2+0*1+6$
$2+0+1+6$
$2*(0-1+6)$
Needs squaring/cubing: $2^2+0+1+6$
$2*(0*1+6)$
$20-1-6$
$20*1-6$
$20+1-6$
$2*0+16$
$2^0+16$
$2+0+16$
$20-1^6$
$20/1^6$
$20+1^6$
$20+\sqrt{\sqrt{16}}$ (thanks @MariaDeleva)
Needs squaring/cubing: $-2+0+(-1+6)^2$
$20+\sqrt{16}$
$20-1+6$
$20*1+6$
$20+1+6$
Needs squaring/cubing: $-2^3+0*1+6^2$
Needs squaring/cubing: $-2^3+0+1+6^2$
Needs squaring/cubing: $(2^2+0+1)*6$
Needs squaring/cubing: $-2^2+0-1+6^2$
$(2+0)*16$
Needs squaring/cubing: $-2^2-0+1+6^2$
Needs squaring/cubing: $-2+0*1+6^2$
Needs squaring/cubing: $-2-0+1+6^2$
$20+16$
Needs squaring/cubing: $2-0-1+6^2$
Needs squaring/cubing: $2+0*1+6^2$
Needs squaring/cubing: $2+0+1+6^2$
$20*\sqrt{\sqrt{16}}$
Needs squaring/cubing: $2^2+0+1+6^2$
Needs squaring/cubing: $(2^3-0-1)*6$
Needs squaring/cubing: $2^3-0-1+6^2$
Needs squaring/cubing: $2^3+0*1+6^2$
Needs squaring/cubing: $2^3+0+1+6^2$
Needs factorial? $2^3+0!+1+6^2$ (thanks @numberknot)
Needs squaring/cubing: $-2-0+(1+6)^2$
Needs squaring/cubing: $(2^3+0*1)*6$
Needs squaring/cubing: $2*0+(1+6)^2$
Needs squaring/cubing: $(2+0)*(-1+6)^2$

edited Apr 13 '17 at 12:50

Community

1

answered Sep 11 '16 at 12:01

Rand al'Thor

116,845
28
322
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That was quick, even if you only got half! – Beastly Gerbil Sep 11 '16 at 12:05
Thank you for these even if its only half ^3^ – NotVeryGood Sep 11 '16 at 12:08
46=2^3+0!+1+6^2 – Numberknot Sep 11 '16 at 13:02
1

@numberknot Is factorial allowed? It's not one of the operations mentioned in the question. – Rand al'Thor Sep 11 '16 at 13:03
@NotVeryGood The list is complete! :-) – Rand al'Thor Sep 11 '16 at 13:33
I'm still hoping to find some way to make 46 without factorial. – Bulldogg6404 Sep 11 '16 at 17:48
@randal'thor 11 can easily be done without square/cube. 2*6-1+0=11 – dcfyj Sep 12 '16 at 12:19
@dcfyj No, because 2,0,1,6 have to appear in the right order. – Rand al'Thor Sep 12 '16 at 12:31
Ah, I wasn't aware of that. – dcfyj Sep 12 '16 at 12:31

score 4 · Answer 2 · edited Apr 13 '17 at 12:50

$2 * 0 + 1 ^ 6$
$2 + 0 * 1 * 6$
$2 + 0 + 1 ^ 6$
$20 - 16$
$2 * 0 + 1 - 6$
$2 * 0 * 1 + 6$
$2 * 0 + 1 + 6$
$2 + 0 * 1 + 6$
$2 + 0 + 1 + 6$
$2 * (0 - 1 + 6)$
$2 ^ 2 + 0 + 1 + 6$
$(2 + (0 * 1)) * 6$
$20 - 1 - 6$
$20 - 1*6$
$20 + 1 - 6$
$2 * 0 + 16$
$2 ^ 0 + 16$
$(2 + 0 + 1) * 6$
$20 - (1 ^ 6)$
$20/1 ^ 6$
$20 + 1 ^ 6$
$20 + \sqrt{\sqrt{16}}$ (Credits Maria Deleva)
$-2 + 0 + (-1 + 6)^2$
$20 + \sqrt{16}$
$20 - 1 + 6$
$20 + 1 * 6$
$20 + 1 + 6$
$(2 ^ 2 + 0) * (1 + 6)$
$-(2 ^ 3) + 0 + 1 + 6 ^ 2$
$(2 ^ 2 + 0 + 1) * 6$
$(2 ^ 2 + 0 + 1)^2 + 6$
$(2 + 0) * 16$
$-(2 + 0 + 1) + 6 ^ 2$
$-2 + 0 * 1 + 6 ^ 2$
$2 * 0 - 1 + 6 ^ 2$
$20 + 16$
$2 + 0 - 1 + 6 ^ 2$
$2 + 0 * 1 + 6 ^ 2$
$2 + 0 + 1 + 6 ^ 2$
$20 * \sqrt{\sqrt{16}}$
$2 ^ 2 + 0 + 1 + 6 ^ 2$
$(2 ^ 3 + 0 - 1) * 6$
$2 ^ 3 + 0 - 1 + 6 ^ 2$
$2 ^ 3 + 0 * 1 + 6 ^ 2$
$2 ^ 3 + 0 + 1 + 6 ^ 2$
$-2 + 0 + (1 + 6)^2$
$(2 ^ 3 + 0 * 1) * 6$
$(2 + 0 - 1 + 6) ^ 2$
$2^0 + (1 + 6)^2$

exponents larger than 3 are not allowed according to the comments — elias, Sep 11 '16 at 12:50

score 2 · Answer 3 · answered Sep 11 '16 at 17:13

2

Some of these become possible with decimal point: $$\begin{matrix} 11=.2^{0-1}+6 \\ 30=.2^{0-1}*6 \\ 35=\dfrac{20+1}{.6} \\ 50=\dfrac{20}{1-.6} \end{matrix}$$

answered Sep 11 '16 at 17:13

Rosie F

8,621
2
18
50

score 1 · Answer 4 · edited Sep 12 '16 at 06:07

1

$17 = (2*0)!+16$

$18 = 2+0+16$

$18 = (2+0+1)*6$

$19 = 2+0!+16$

$20 = (2+0!+1)!/6$

$21 = (2+0!)(1+6)$

$22 = (2+0!)!+16$

$23 = 2^3-0!+16$

$24 = (2+0!+1)*6$

$25 = (2-0-1-6)^2$

You should be able to continue on your own.

edited Sep 12 '16 at 06:07

EKons

1,253
12
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answered Sep 11 '16 at 12:12

Machining Machine

11
2

7

what's the actual pattern that allows us to continue on our own? – elias Sep 11 '16 at 12:47
When I answered the question it was different. I was just giving a little hint since he had just gotten stuck at #17. I stopped because it's his homework and it's tedious work. – Machining Machine Sep 11 '16 at 12:51

Making 1-50 from 2016

4 Answers4

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