Please do **NOT** post your answer, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

Two quick puzzles this week.

1) Can you use three 2s to create an expression (such as ‘2+2+2’) that equals 11?

2) Can you use each of the digits 0-9 once to create an expression that equals 1?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle **(UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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The second one took me about a minute, with a pen and a post-it note. I can’t do the first one though. I bet it’s dead obvious but I can’t do it. I even tried to cheat and I googled it and found a website posing a similar question — use two number 2s and any symbol to make 5 — and you wouldn’t believe the arguments!

Just commenting at the top to warn you if you worry about such things, that I posted an answer to Q1 below, and others have posted answers to Q2. It really doesn’t matter that Prof Wiseman says not to post the answer.

I’ve done the first one using a bitwise operator. Not sure if this is what’s intended.

1.) three twos or ONLY three twos? 🙂

2.) endless solutions

I really hope it’s ONLY three 2s or I’ve been wasting my time! If you can use as many as you want then it’s a piece of piss.

Well said, Anne.

0.1s

1) no

2) no

Actually my favourite comment ever.

Actually when Richard says “can you…”, I think he means “can one…”

You think wrong, Edster.

I can only do the first puzzle if I tri to base my maths along on different thinking.

Both simple done quickly. Ask me how.

Yes and no. 😉

How?

2+2+2+16

I satisfied the criteria of using three 2s and there is nothing in the wording that suggests that I cannot use other numbers.

35 148

— + — = 1

70 296

Classic spoiler, wee man. Good work… The only problem is, your first answer’s wrong.

If is good to have a puzzle that does not wrap the question up in ambiguous words. (Such as “I have three couples coming to a party and I need to seat them at a table in 22 ways. What should I do?”)

If factorials are allowed we can have (2! + 2!)! – 2!

Thanks Barry – you are a star

Log (22)^2

In your rush to be smug, you over complicated the answer.

(2+2)! – 2 = 22

sqrt(22^2) also works.

You’re right

(2+2)! – 2

I appreciate a spoiler who can learn from his mistakes. Sterling work, wee man.

Woah, serious brain fart. Here’s what I should have typed: sqrt(22)^2

One minute for the two.

How long for each?

the second puzzle has an embarrassingly simple solution. staywithusscotland’s answer is nice too.

since the easiest answer to the first puzzle has already been spoiled, i’m curious how many solutions work.

Off the top of my head

138/276 + 45/90 = 1

140/368 + 57/92 = 1

143/528 + 70/96 = 1

145/290 + 38/76 = 1

148/296 + 35/70 = 1

169/507 + 32/48 = 1

185/370 + 46/92 = 1

186/372 + 45/90 = 1

204/867 + 39/51 = 1

207/549 + 38/61 = 1

208/793 + 45/61 = 1

231/609 + 54/87 = 1

269/807 + 34/51 = 1

273/406 + 19/58 = 1

276/345 + 18/90 = 1

284/710 + 39/65 = 1

287/369 + 10/45 = 1

306/459 + 27/81 = 1

307/614 + 29/58 = 1

307/921 + 56/84 = 1

308/462 + 19/57 = 1

309/618 + 27/54 = 1

310/465 + 29/87 = 1

315/609 + 42/87 = 1

351/702 + 48/96 = 1

357/408 + 12/96 = 1

369/574 + 10/28 = 1

372/465 + 18/90 = 1

375/480 + 21/96 = 1

381/762 + 45/90 = 1

405/729 + 36/81 = 1

417/695 + 32/80 = 1

426/710 + 38/95 = 1

451/902 + 38/76 = 1

473/528 + 10/96 = 1

481/962 + 35/70 = 1

485/970 + 13/26 = 1

485/970 + 16/32 = 1

485/970 + 31/62 = 1

486/972 + 15/30 = 1

504/623 + 17/89 = 1

507/819 + 24/63 = 1

540/972 + 36/81 = 1

609/783 + 12/54 = 1

630/945 + 27/81 = 1

678/904 + 13/52 = 1

728/936 + 10/45 = 1

735/840 + 12/96 = 1

748/935 + 12/60 = 1

Very funny, Sheldon! That made me laugh out loud.

It’s like all of my birthdays and Christmases have come at once.

PS You must have a very big head.

I don’t have an answer for the first yet (except for the spoiler I just read). But I’ve got 362880 simple answers for the second, but there are many many more answers possible!

Second one is so simple – a very large number of possibles (can’t be bothered to work out how many) but they bear no resemblance to the ones given above!

1) I’d tried powers, but until I looked at the comments, I didn’t think of using factorials.

2) As others have said, there are numerous solutions, but among them is at least one which only uses addition and subtraction, plus uses all the digits in order without combining them (e.g. 5 + 6 + 7 + 8, not 56 + 78).

Since the cat is out of the bag ((0/9)+1+2+7+8)/(3+4+5+6)=1

Why go to so much trouble?

0*(2+3+4+5+6+7+8+9) + 1

If we’re allowed to spoil, for 2), put the numbers 1-9 in a bracket doing anything you want, then raise the bracket to the power of zero.

this. i’m surprised so many people went for the more complicated arithmetic solutions first.

Man rock it old school, blood. None them factorial ting, allow it.

9-8+7-6-5+4+3-2-1 = 1

@Gareth. You answer fails at the first hurdle as it does not use the zero.

I have made an answer that looks like your but does use the zero:

(9 + 8 + 7 + 6 + 5 + 4 + 3 + 2) x 0 + 1 = 1

It is possible to create some variations on that answer by changing some of the pluses to minuses or multiplies. Some also can be changed to divides but take care if dividing by zero. It is also possible to change all the numbers except the zero to number factorial!

Many other similar changes can be made. I think this answer is one of the most flexible templates possible for this puzzle.

Allow it, Bazza! Point blank scerrrrutch!

9-8+7-6-5+4+3-2-1+0 = 1

Bare cheers bludfam.

|√(22^2)| is a valid alternative solution to #1.

Best answer.

My answer to the second was (123,456,789)^0 = 1.

Also (0-1)+(2-3)-(4-5)-(6-7)-(8-9) = 1

Got both fairly quickly. The 0-9 one took a bit longer and involved pen and paper (well, keyboard and screen, to be precise). 🙂

Many answers have already been given… so I guess I’m not spoiling anything with mine:

1. (2+2)! – 2 (already given above)

2. (9-8) x (7-6) x (5-4) x (3-2) x (1-0) = 1

I also like Eddie’s answer of raising digits 1-9 to the 0th power.

I like your Number 2, Ken.

Me 2.

#1. (2 * 2)! – 2 = 22

#2. 123456789^0=1

Today is day when everybody breaks rule. lol

The dog travelled 10 km. Sorry, I’m late with the answer

First one took some time, during which I also found a way to get 178000 billions using only three 2’s and the sinus function.

Second one took one try (adding and subtracting the numbers until they fit).

I haven’t heard of that – who nose what it is?

I couldn’t do the first until I read the above but I got the same answer as Ken in a few seconds

Nobody seems to care about not spoiling this one. But there are a large number of answers to both questions. I got one answer for each puzzle in about thirty seconds, but there are obviously more.

Q1 took me ages, but I got there and kicked myself for taking so long. Q2 took about 2 mins.

To complete the posting of solutions we should also note:

CEILING [2 22]

FLOOR [1 234567890]

Where (if we had the ability to enter mathematical cymbals) CEILING and FLOOR represent the cymbals for “find the set element that is smallest/largest”.

We could for example use this to get 22 from any number of 2s. This is probably an optimal solution to the puzzle.

Please note that not all the posts from me have been from me. Once again someone else (perhaps more than one someone) has used my name to cover up their own name in disguise.

what have cymbals got to do with it? That answer is just noise

Wow; talk about self-realization starvation…

(1) about 1 second

(2) about 20 seconds – 15 seconds before I noticed an easy way to do it, then 5 seconds to check.

I wrote my answer (above) before reading the thread. As so many people have already given their answers, I might as well do so too:

(1) (2+2)! – 2

(2) (9+8+6) – (7+5+4+3+2+1) = 1

I did number (2) by starting with the fact that 9+8+7+6+5+4+3+2+1 = 45, so it was just a matter of splitting it into one lot of 23 and another lot of 22.

But I like the (anything)^0 option. How about (9^(8^(7^(6^(5^(4^(3^(2+1)!)!)!)!)!)!) ^ 0 ?

With the way many people reply, I’d suggest Richard should just change the rules, inviting people to post their answers right away. Tell everybody not to read the comments if they don’t want to see spoilers.

I wonder if Richard even knows that people regularly post the answers – there’s no evidence that he reads the comments.

These two questions have us all completely stumped.

Would anyone consider giving us a teensy weeny clue?

If the three of you glance at some of the other comments you will find a range of possible answers, some with explanations of why they are right and some with explanations about why they are wrong.

These answers may not correspond with the official answer which will be available on Monday (local time). The official answer is usually given without further explanation and that off times leads to discussions about precise readings of the original puzzled necessary to arrive at the official solution.

’2+2+2′ = 2

0-9 = 1

wottatwit

’2+2+2′ = 22

of course

Since almost everyone here ignore the rules…

1: abs(22.2)

2: (any combination)^0

I think you mean floor(22.2) or round(22.2). abs(22.2) is still 22.2.

Yes, I can do both. The first involved signs and the second factorials.