+0

# Problem

+5
623
9
+1068

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?

Apr 30, 2015

#5
+95361
+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

May 1, 2015

#1
+95361
0

I know how to show this

May 1, 2015
#2
+1068
+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.

May 1, 2015
#3
+95361
0

Do you want me to show you?

Maybe you want a little hint?

May 1, 2015
#4
+1068
0

May 1, 2015
#5
+95361
+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
#6
+4682
+5

Wow Melody.

Cool use of Algebra!

May 1, 2015
#7
+95361
+5

Yep - algebra is really cool

And later you will find that Calculus is EVEN cooler !!

May 1, 2015
#8
+1068
+5

Thanks, Melody! It is interesting.

May 1, 2015
#9
+4682
+5

O.O

i wanna learn Calculus!

May 1, 2015