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## Question

How can you get five out of four?

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always do the extra credit question

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"Hey guys! could you please lend me 5 bucks??"

Guy 1: "only got 2 bucks."

Guy 2: "I can throw in an extra buck."

Guy 3: "I'll let you borrow 1.50"

Guy 4: "Here's the extra 0.5 that you need."

I got 5 (bucks) out of 4 (guys).

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"Hey guys! could you please lend me 5 bucks??"

Guy 1: "only got 2 bucks."

Guy 2: "I can throw in an extra buck."

Guy 3: "I'll let you borrow 1.50"

Guy 4: "Here's the extra 0.5 that you need."

I got 5 (bucks) out of 4 (guys).

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IV minus the "I". IV=4 V=5

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\ /

\/

(4 slashes making 5)

A family of 4 with a pregnant mom will soon be 5

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have someome give you an extra/always have a spare?

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How can you get five out of four?

f. i. v. e.

five out of 4 letters.

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Kinda vague...

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Yes

Edited by Molly Mae
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4= IV. Take away the I and you get V (V=5)

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edit for double post

Edited by Aaryan
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Here's a four:

/|
¯|¯

Take the five line segments which make this figure

and arrange them this way:

_
|
¯|
¯

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f. i. v. e.

five out of 4 letters.

well done..

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Here's a four:

```
/|

¯|¯

```
Take the five line segments which make this figure and arrange them this way:
```
_

|

¯|

¯

```

nice...

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4 = (2 + 2)

define a = b = 2, so

4 = (a + b)

4 = (a + b) x 1

4 = (a + b) x (1 / 1)

4 = (a + b) x [ (c - d) / (c - d) ]

note that this would hold regardless of the value of c and d, but to make things simple I'll also make c = d = 2

4 = (a + b) x (c - d) / (c - d)

4 = [ a(c - d) + b(c - d) ] / (c - d)

4 = [ a(c - d) + b(c - d) + 0 ] / (c - d)

4 = [ a(c - d) + b(c - d) + (c - c)] / (c - d)

since c = d we can substitute the last c in the numerator with a d

4 = [ a(c - d) + b(c - d) + (c - d)] / (c - d)

4 = [ (a + b + 1) x (c - d) ] / (c - d)

4 = (a + b + 1) x [ (c - d) / (c - d) ]

4 = (a + b + 1) x [ 1 / 1 ]

4 = a + b + 1

4 = 2 + 2 + 1

4 = 5

edit: put everything in a spoiler, although I'm not exactly sure why given the nature of this riddle

Edited by plasmid
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4 = (2 + 2)

define a = b = 2, so

4 = (a + b)

4 = (a + b) x 1

4 = (a + b) x (1 / 1)

4 = (a + b) x [ (c - d) / (c - d) ]

note that this would hold regardless of the value of c and d, but to make things simple I'll also make c = d = 2

4 = (a + b) x (c - d) / (c - d)

4 = [ a(c - d) + b(c - d) ] / (c - d)

4 = [ a(c - d) + b(c - d) + 0 ] / (c - d)

4 = [ a(c - d) + b(c - d) + (c - c)] / (c - d)

since c = d we can substitute the last c in the numerator with a d

4 = [ a(c - d) + b(c - d) + (c - d)] / (c - d)

4 = [ (a + b + 1) x (c - d) ] / (c - d)

4 = (a + b + 1) x [ (c - d) / (c - d) ]

4 = (a + b + 1) x [ 1 / 1 ]

4 = a + b + 1

4 = 2 + 2 + 1

4 = 5

edit: put everything in a spoiler, although I'm not exactly sure why given the nature of this riddle

as:

c=d=2

so

c-d=0

and you can not devide on 0

Edited by wolfgang

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