2013 is the first year in over two decades where each digit is different!
Image: Thinkstock
It’s 2013, and to celebrate the first issue of Maths and Stats by Email for the new year, we’ve got a puzzling problem for you to solve!
Using the digits 2, 0, 1, 3 once each and any of the signs + – x ÷, can you make the number 1? How about 2? Can you make each of the numbers 1–10? You might need brackets () so you can do addition and subtraction before multiplication and division.
There’s a heap of solutions for you to find. Bonus marks for using fewer signs (but still using all of the digits), not using brackets and for keeping the digits in the original order: 2, 0, 1, 3. Leave your best answers in the comments – we’ll be posting our attempt in a few days.
15 January 2013 at 10.28 pm
(1*0)+3-2=1
3-1+(2*0)=2
3+((1-2)*0)=3
3+1-(2*0)=4
3+2-(1*0)=5
3*2+(0/1)=6
3*2+1-0=7
(3+1)*2-0=8
(2+1)*3-0=9
30/(1+2)=10
16 January 2013 at 7.29 am
2013 is the first year to have four different digits since 1987, to be exact
My answers are:
-2+(0×1)+3=1
-2+0+1+3=2
2x0x1+3=3
2×0+1+3=4
2+0+(1×3)=5
2+0+1+3=6
(2+0)x3+1=7
(2+0)x(1+3)=8
(2+0+1)x3=9
-20/(1-3)=10
23 January 2013 at 11.27 am
I love your answer!
The digits are in order for every answer except 7. I thought about it for a bit – how about:
(20+1)/3=7 ?
24 January 2013 at 8.43 am
I tried to make all the digits in order. Seven stumped me, so I made it in order as much as I could. I never thought about division, though… I like your seven!
23 January 2013 at 6.48 pm
very good question i loved it…
My answers is:
-2+(0×1)+3=1
-2+0+1+3=2
2x0x1+3=3
2×0+1+3=4
2+0+(1×3)=5
2+0+1+3=6
(20+1)/3=7
(2+0)x(1+3)=8
(2+0+1)x3=9
-20/(1-3)=10
24 January 2013 at 6.06 pm
You copied me and David!
25 January 2013 at 10.18 am
That’s a really interesting thought. The answer is still correct, whether it was copied or come up with independently. For example, there are correct answers for many school maths tests. So, you and your classmates may very well have the same answers even if you don’t copy each other. When the answers are correct, how can you prove someone has copied?
25 January 2013 at 10.25 am
Anyone up for trying it with 2013 backwards, i.e. 3102? I need to get back to work, but here’s a start:
3+(1×0)-2=1
3+1-0-2=2
3+(1x0x2)=3
28 January 2013 at 6.25 am
Well, I believe it might be copied, as he could have easily come here to show he’s answers, then see mine (in order) and David’s seven in order. Unfortunately, I came here to show my answers the moment I got my email. I had none of them in order… Zero was at the start of all the equations. When I saw that you needed them in order (2,0,1,3) I changed them all (mostly)
28 January 2013 at 7.00 am
This is an interesting question… How about for number 11 and up, you can use the numbers twice, i.e. 31023102?
(31/0)+2=2
3+1+(0×2)=4
3+(1×0)+2=5
3+1+0+2=6
3
(3+1+0)x2=8
3x(1+0+2)=9
3
3+1+0+(2×3)+1+(0×2)=11
3+1+0+2+3+1+0+2=12
((3+1+0)x2)+3+(1×0)+2=13
((3+1+0)x2)+3+1+0+2=14
3+(1×0)x(2+3)+(1x0x2)=15
(3+1+(0×2))x(3+1+(0×2))=16
This is it so far, I’ll come back and do more