a simple brain teaser

keithsfleet

Use each number once and once only..
Each number must be used.

2, 3, 4, 5 = 28 how?

Don't post the answer if you get it.
Pm me if you want it

Micky Dolenz

Easy peesy.

shopaholic01

keithsfleet wrote: »

Use each number once and once only..
Each number must be used.

2, 3, 4, 5 = 28 how?

Don't post the answer if you get it.
Pm me if you want it

You've just killed your own thread!

the keen edge

shopaholic01 wrote: »

You've just killed your own thread!

Ah stop, sure there is so much to discuss.

keithsfleet

shopaholic01 wrote: »

You've just killed your own thread!

Ha yea sorry a few pre pub drinks in me.
The head ain't the clearest.
Ok people, fire away if you want to answer it.

EDDIE WATERS

two cubed+4 times 5 =28

NuckingFacker

I never was much good at maths...brain hurts...

Ficheall

People answering will still kill the thread.

There are multiple solutions - perhaps you could define what operations you're suggesting people use?

Ficheall

EDDIE WATERS wrote: »

3²+4 times 5 =28

WRONG! :P

artanevilla

keithsfleet wrote: »

Use each number once and once only..
Each number must be used.

2, 3, 4, 5 = 28 how?

Don't post the answer if you get it.
Pm me if you want it

13spanner

(4 x 5) + 2 cubed (which would be a lil' 3) = 28.

keithsfleet

It's incredibly simple.
You don't need to Square anything.

Purely use addition multiplication and subtraction.

sock puppet

Is this from the Special Olympics version of Countdown?

tritium

EDDIE WATERS wrote: »

3²+4 times 5 =28

I think you mean two cubed rather than three squared

Skid X

keithsfleet wrote: »

Use each number once and once only..
Each number must be used.

2, 3, 4, 5 = 28 how?

Don't post the answer if you get it.
Pm me if you want it

((2*5)-3)*4 = 28

keithsfleet

Skid X wrote: »

((2*5)-3)*4 = 28

Congrats you got it

Nailz

Ficheall wrote: »

People answering will still kill the thread.

There are multiple solutions - perhaps you could define what operations you're suggesting people use?

This has become unfair just by you being here. :P

humbert

No cubing required.

not yet

I don't know what a tracker mortgage is...

tritium

humbert wrote: »

No cubing required.

But a perfectly valid (and slightly flashy ) approach

not yet

not yet wrote: »

I don't know what a tracker mortgage is...

Well, that sums up this thread!

kraggy

Take the diagonal part of the number 4 away and it becomes a plus sign.

Then you don't even have to change the order of the numbers or do anything else and it's right.

2 3 + 5 = 28

I love countdown

Ficheall

bluewolf wrote: »

I love countdown

We need a Rachel.

Lawrence1895

Can I send a PM, that I hate maths?

shopaholic01

Ficheall wrote: »

We need a Rachel.

And a few consonants and vowels.

Ficheall

Ficheall wrote: »

We need a Rachel.

Hasn't been the same since whiteley and vorderman!

kraggy

Here's one while we're at it.

You have

a. A weighing balance
b. 9 balls

8 of the balls are the same weight, one is heavier than the others.

Question: What is the least number of times you can think of using the weighing balance to determine which ball is of a heavier weight?

Ficheall

bluewolf wrote: »

Hasn't been the same since whiteley and vorderman!

Rachel and Susie are still there and that's good enough for me.

Chuck Stone

2 pears + 3 apples + 2 peaches

=

A fruit salad

kraggy

kraggy wrote: »

Here's one while we're at it.

You have

a. A weighing balance
b. 9 balls

8 of the balls are the same weight, one is a different weight.

Question: What is the least number of times using the weighing balance necessary to determine which ball is of a different weight?

(Who can write the question in better English than that please?)

I originally thought 5, but you can probably group them together and do it fewer times
2 or 3

Ficheall

kraggy wrote: »

Here's one while we're at it.

You have

a. A weighing balance
b. 9 balls

8 of the balls are the same weight, one is a different weight.

Question: What is the least number of times you can think of using the weighing balance to determine which ball is of a different weight?

3

Do I get a prize for beating bluewolf?


Edit: I presume it's a lower bound on the maximum number of necessary weighings to determine which ball is of a different weight that you are looking for.

kraggy

Ye can't just guess a shaggin number! Ye have to explain yourselves!!

Ficheall

Ficheall wrote: »

3

Do I get a prize for beating bluewolf?

Hey, that was one of my answers :pac::pac::pac:

sock puppet

It's 2.

Ficheall

kraggy wrote: »

Ye can't just guess a shaggin number! Ye have to explain yourselves!!

And spoil it for others? Don't be daft.
It's a gorgeous problem.

Micky Dolenz

3, has to be.

Give us something hard, oh er matron.

Skid X

8 out of 10 Cats does Countdown is on Channel 4 at 9pm.

Featuring Rachel Riley

kraggy

Micky Dolenz wrote: »

3, has to be.

Give us something hard, oh er matron.

nope.

Vowel please, carol

kraggy

bluewolf wrote: »

Vowel please, carol

OOOOOoooooo Carol...

Ficheall

kraggy wrote: »

nope.

What do you mean, 'nope'? It is 3.

kraggy

Ficheall wrote: »

What do you mean, 'nope'? It is 3.

It's not.

sock puppet

sock puppet wrote: »

It's 2.

kraggy wrote: »

Ye can't just guess a shaggin number! Ye have to explain yourselves!!

Fine. Weigh 6 and leave 3 to the side. If they don't balance weigh 2 of the 3 on the heavy side and find the heavy one. If they do balance weigh 2 of the 3 remaining balls.

I've rather gone off these problems since employers ask them in interviews.

Ficheall

Ficheall wrote: »

Unless it's a trick question, or you haven't given us enough information, such as there being a limit of the number of balls that fit on the balance, then it is.

You didn't beat me! Hand back the prize!

kraggy

sock puppet wrote: »

Fine. Weigh 6 and leave 3 to the side. If they don't balance weigh 2 of the 3 on the heavy side and find the heavy one. If they do balance weigh 2 of the 3 remaining balls.

I've rather gone off these problems since employers ask them in interviews.

Correct

Ficheall

kraggy wrote: »

Correct

No, it isn't.

kraggy

Ficheall wrote: »

Unless it's a trick question, or you haven't given us enough information, such as there being a limit of the number of balls that fit on the balance, then it is.

You only need to use the balance 2 times. As demonstrated by my lovely assistant Sock.

Ficheall

kraggy wrote: »

You only need to use the balance 2 times. As demonstrated by my lovely assistant Sock.

No, I'm afraid you're mistaken. You might want to re-read your question.

kraggy

Ficheall wrote: »

No, it isn't.

Yes it is. The wording might be different to how I'd put it, but the essence is the same.

humbert

I was trying to think how I'd do it with my electric scales