Numbers divisible by 9

[JimDom]

How does this always work??

1: Take any number 10 or greater.

2: Take the sum of the digits and subtract them from the original number to get a new number.

3: That new number is always evenly divisible by 9 with no remainder.

Two examples:
5826
5 + 8 + 2 + 6 = 21
5826 - 21 = 5805
5805 / 9 = 645

1358280
1 + 3 + 5 +8 +2 + 8 + 0 = 27
1358280 - 27 = 1358253
1358253 / 9 = 150917

It works for any number!

 

[Deleted member 4993]

How does this always work??

1: Take any number 10 or greater.

2: Take the sum of the digits and subtract them from the original number to get a new number.

3: That new number is always evenly divisible by 9 with no remainder.

Two examples:
5826
5 + 8 + 2 + 6 = 21
5826 - 21 = 5805
5805 / 9 = 645

1358280
1 + 3 + 5 +8 +2 + 8 + 0 = 27
1358280 - 27 = 1358253
1358253 / 9 = 150917

It works for any number!

Suppose you have a 3 digit number XYZ

Numerical value of XYZ is:

100*X + 10*Y + Z ................(1)

(100*X + 10*Y + Z) - (X + Y +Z)

= 99 * X + 9 * Y ..................(2)

Do you see that (2) is divisible by 9?

This process can be extended to any number of digits.

 

[Steven G]

How does this always work??

1: Take any number 10 or greater.

2: Take the sum of the digits and subtract them from the original number to get a new number.

3: That new number is always evenly divisible by 9 with no remainder.

Two examples:
5826
5 + 8 + 2 + 6 = 21
5826 - 21 = 5805
5805 / 9 = 645

1358280
1 + 3 + 5 +8 +2 + 8 + 0 = 27
1358280 - 27 = 1358253
1358253 / 9 = 150917

It works for any number!

You can not conclude that something is true by just checking a few cases. You can either check that this is true all numbers--which is not possible in this case since there are an infinite number of numbers- or you can prove the conjecture for any positive integer as Dr Khan did in the post above for 3 digit numbers.

 

[JimDom]

Suppose you have a 3 digit number XYZ

Numerical value of XYZ is:

100*X + 10*Y + Z ................(1)

(100*X + 10*Y + Z) - (X + Y +Z)

= 99 * X + 9 * Y ..................(2)

Do you see that (2) is divisible by 9?

This process can be extended to any number of digits.

Thank you Subhotosh! I understand now.