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## you are

## 584 is possible

## 5/5 = 1

## han gon

## Got 589

Will keep trying

## 580 now

Bit closer

## Is this even possible?

## don't think so

## dunno, i randomly generated them

## Thought I had it then.

But then had to make 8 out of 5,5 and 6. Made 7.

## 580

grrr

## 584

don't think I can get it

## (75+25-1)x6 = 594

594-5-5 = 584

## this is as close as i got

## very good

i did this then my boss wanted me to show him how to do an hlookup. quite annoyed! although i do love an hlookup

## Didnt get this far. Would probably have done given time

I was trying to get to 608. I worked out that (75+1) * 8 - 25 = 583. Couldnt make the 8 though.

## aye.

## 1*5*6*5*25*75

281250.

Way off.

## do they get a calculator on countdown btw?

## i don't think so

## you sunk my battleship!

## Yeah, 584 also.

*shows pad to Gyles Brandreth*

## Great, a LOST thread!

## yeah, think i've got it.

## 1258

## if one of the 5s was a 4 it'd be a piece of piss

(5+6) x (75-25+4-1)

## or if one of the 5s was a 6

## or that

but a prime factor solution is more elegant, I'm sure you'll agree ;)

## it isn't

## you are a harsh mistress, c_r

## your victories are worth nothing if you don't accept your defeats

## (5-1)^5 - (75 x 6) = 574

25 = 9

574 + 9 = 583

## 25=9?

## all square numbers are same

its a squa re, same shape

## has cat_race hacked your account??

## all account

## (5-1)^(25/6) + 75 - 5 = 582

this is the most badass answer anyone's going to give

It's 100% impossible.

The deviation of 583 from a multiple of 5 will always be of the form

5n-3 (1)

5n-2 (2)

where n is a positive integer, depending on whether it is too big or two large. As the only non-5 numbers to choose from are 6 and 1 we can't form numbers that satisfy (1) and (2) and therefore the question has no real solutions. QED

## get fucked cunty

## this is why DiS is going to the dogs

nonetheless I wonder if Grouchy's answer, intelligently as it is argued, could be affected by the use of indices, as well as divisions of different multiples of 5...obviously anything involving 5n^x will result in 5m, but perhaps x^5n could be combined with y or y^5m, or even (5l/5k)^5j...the numbers being raised to powers would have to share factors. For example, 75/25 is 3; 3^5 is 243 and 6^5 is 7776; the divisor is 32, which when added to 1 becomes 33, which is 5n+3/5n-2...you see, in theory, how 583 could be attained

aye but you can't just throw the rules out can you. you could make the same argument for any arbitrary function which would make the game trivial

## are indices not allowed on countdown?

bullshit! it's called bidmas for a reason

## anyway, 75/25 is 3

there's your 3. 5n/5m = n/m = l