Use four 0's & any mathematical operators to get 24

@thebigk • Oct 26, 2024

Oct 26, 2024

1.8K

For those who don't have to do anything better -

Use four 0's (zeros) and any mathematical operators to get 24.

Those who are smart, don't spoil the fun soon. Others, give it a try.

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raj87verma88

@raj87verma88-ZpL2Wn • May 2, 2008

Rather difficult
But let me give it a shot

cos-1 (inverse) 0 = 90
√90 = 9.5
and 3√90 = 4.5
√cos (inverse) 0 + √cos (inverse) 0 + √cos (inverse) 0 -3√cos (inverse) 0
= √90 + √90 + √90 - 3√90
= 9.48 +9.48 + 9.48 – 4.48
= 28.44 – 4.48
= 23.9

after taking square root and cube root if we approximate

that is

9.5 + 9.5 + 9.5 – 4.5
then answer is exactly 24

How about it Biggie
0
raj87verma88

@raj87verma88-ZpL2Wn • May 2, 2008

Also instead of cube root of 90 we can do
ln90 = 4.49
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Kaustubh Katdare

@thebigk • May 2, 2008

Hint -

K.I.S.S 😁

...and since when did we start relying on 'approximate' values?

On the same lines, I've another question -> is 0.999999...(infinite times) equal to 1 ?

If yes, how? If not, why? 😉
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devesh

@devesh-f6jAzZ • May 4, 2008

do we only have to use 0's?
I mean nothing else(no other numerics),only operators and zeroes!!!
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Kaustubh Katdare

@thebigk • May 4, 2008

Yes, that's right. Only four zeros! (and of course, mathematical operators).

Is it really 'that' difficult?
0
gohm

@gohm-F3UUpP • May 4, 2008

Are we allowed to rotate the operators?
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gohm

@gohm-F3UUpP • May 4, 2008

Nope, it is never equal to 1. Only 1 is equal to 1. In most practical uses it is close enough mathematically to round the .99999 to 1 however it is never actually equal to 1.

The_Big_K
Hint -

K.I.S.S 😁

...and since when did we start relying on 'approximate' values?

On the same lines, I've another question -> is 0.999999...(infinite times) equal to 1 ?

If yes, how? If not, why? 😉

0
Kaustubh Katdare

@thebigk • May 4, 2008

1. No, you cannot 'rotate' operators. Come on, its just basic mathematics!

2. If 0.99999...(infinite) is not equal to one, what's wrong with the following? -

x = 0.99999...(infinite times)
10x = 9.9999.......

=> 9x = 9

=> x = 1

😁

What say? 😉
0
gohm

@gohm-F3UUpP • May 4, 2008

Here is your problem, the tricky business between the 10x=9.999999... and then 9x=9. Did you change the value of x?

The_Big_K
1. No, you cannot 'rotate' operators. Come on, its just basic mathematics!

2. If 0.99999...(infinite) is not equal to one, what's wrong with the following? -

x = 0.99999...(infinite times)
10x = 9.9999.......

=> 9x = 9

=> x = 1

😁

What say? 😉

0
Kaustubh Katdare

@thebigk • May 4, 2008

gohm
Here is your problem, the tricky business between the 10x=9.999999... and then 9x=9. Did you change the value of x?

Umm, okay, here we go -

x = 0.99999...(infinite times) - Equation (A)
10x = 9.9999....... - Equation (B)

Equation (B) minus (A)

=> 9x = 9

=> x = 1

👍
0
mahul

@mahul-ZxpiLA • May 4, 2008

Four zeroes to score 24, that's easy-->>

(0!+0!+0!+0!)!=24

i hope that would do, unless _k has a problem with my repeated use of factorials 😀
0
Kaustubh Katdare

@thebigk • May 4, 2008

mahul
Four zeroes to score 24, that's easy-->>

(0!+0!+0!+0!)!=24

i hope that would do, unless _k has a problem with my repeated use of factorials 😀
Hah, no problem with factorials! That's the right way to get 24. Pretty easy, eh? 😀
0