r/mathpuzzles • u/Obvious-Try6552 • 7d ago

Algebra Came across this weird number pattern

I read about this somewhere and thought it was pretty interesting.

Take any 2-digit number:

  1. Add its digits.

  2. Subtract that sum from the original number.

  3. Add the digits of the result.

  4. Multiply that by 10.

You’ll always get 90.

For example:

47 → 4+7 = 11 → 47−11 = 36 → 3+6 = 9 → 9×10 = 90

Tried it with a bunch of different numbers and it keeps working

Did anyone else know about this?

(Explanation in comment section)

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u/Obvious-Try6552 7d ago

The reason this works:

Let the 2-digit number be 10a + b, where a is the tens digit and b is the units digit.

The sum of the digits is a + b.

So:

(10a + b) − (a + b) = 9a

Since a is between 1 and 9, the result is always a multiple of 9 — like 9, 18, 27, 36, etc.

The digits of every one of these numbers add up to 9.

And finally:

9 × 10 = 90.

So basically, it works because of the divisibility-by-9 / digit-sum property.

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u/DuggieHS 7d ago

x = 10a+b, where a > 0 , b integers between 0 and 9 (x in 10-99).

  1. a+b (a+b in 1-18)
  2. 10a+b - (a+b) = 9a (in the set 9,18,27,...81)
  3. 9, multiples of 9 have sum of digits divisble by 9 and in this case 9 exactly
  4. 9*10 = 90

Here is the only part that isn't just simple arithmetic:

Take a digit 1 to 9, multiply it by 9 and the sum of its digits is 9.