# Ancient Math Riddle Take 9 from 6, 10 from 9, 50 from 40 and leave 6.

How Come ?

### 6 guesses to Ancient Math Riddle

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• Crabman

Using some funny binary maybe?

First, binary/base 10 equivalents
110 = 6
1001 = 9
1010 = 10
101000 = 40
110010 = 50

If there are the same # of 1’s in the two numbers the answer is 1, if there are a different amount of ones, the answer is zero. I realize this is a stretch, but the last number riddle counted closed areas, multiplied by digets without closed areas, so I’m just taking a shot. Only thing I could come up with.

110 – 1001 = 1 (same # of 1s)
1001 – 1010 = 1 (same # of 1s)
101000 – 110010 = 0 (different # of 1s)

Combine these and you get 110, convert from binary to base10 and you get the 6.

• Dude

Sorry Crabman, that was not what I was looking for.
That ROMAN way of looking at numbers should keep you trying.

• Frank

VI – IX = V-X
IX – X = I
XL – L = X

V-X + I – X = VI = 6

• Frank

Should be (changed – to +)
VI – IX = V-X
IX – X = I
XL – L = X

V-X + I + X = VI = 6

• Achu

Use roman numerals:
Remove nine(IX) from SIX = SIX – IX = S
Take 10(X) from 9(IX) = IX – X = I
Take 50(L) from 40(XL) = XL – L = X
So the answer would be SIX. (6)

• Dude

Achu, you are today’s winner!
SIX – 9 (IX) = S. 9 (IX) – 10 (X) = I. 40 (XL) – 50 (L) = X => SIX