Take away from your year of birth the sum of the four numerals that make it up. You will end of with a number divisible by 9. Why is this?

civonamzuk
Apr 30, 2015

#5**+8 **

Your birthyear has 4 digits let it be abcd

I mean

1000a+100b+10c+d that is the year you were born.

now if I add the 4 digits i get a+b+c+d

so doing the subtraction I get

1000a+100b+10c+d -(a+b+c+d)

=1000a+100b+10c+d -a-b-c-d

=999a+99b+9c

=9(111a+11b+c)

Which is obviously divisable by 9 since 9 is one of the factors

Isn't mathematics fun

Melody
May 1, 2015

#2**+5 **

The sum of numbers from 0 ==> 9 is divisible by 9;

0+1+2+3+4+5+6+7+8+9 = 45 45 / 9 = 5

Let's take year 1776;

1776-21=1755 1755/9=195

This is true for all multi digit numbers starting with # 10;

10-1=9 9/9=1

78-15=63 63/9=7

358-16=342 342/9=38

486,372,103 - 34 = 486,372,069 486,372,069 / 9 = 54,041,341

The question is: Why is this so?

I don't have the right answer to this question, but what I see from these examples is that the #9 has a 'special' place in a structure of multi digit numbers.

civonamzuk
May 1, 2015

#5**+8 **

Best Answer

Your birthyear has 4 digits let it be abcd

I mean

1000a+100b+10c+d that is the year you were born.

now if I add the 4 digits i get a+b+c+d

so doing the subtraction I get

1000a+100b+10c+d -(a+b+c+d)

=1000a+100b+10c+d -a-b-c-d

=999a+99b+9c

=9(111a+11b+c)

Which is obviously divisable by 9 since 9 is one of the factors

Isn't mathematics fun

Melody
May 1, 2015