Why does the number 9 when multiplied by any number and all of the digits are added together leave us with 9?

Question

5 x 9 = 45 - 4+5 = 9,
60 x 9 = 540 - 5 +4 +0 = 9 etc, etc

Answer 1

Try multiplying by 11...9x11=99; 9+9=18. Of course, if you then do 1+8=9, you do get down to 9 eventually. This is because 10=1 (mod 9), so adding digits preserves divisibility by 9.

Answer 2

yes this is a special to the number '9'
as some ppl proved that it is wrong in some cases but it is not true
9 * 11 = 99
9+ 9 = 18 again 1+ 8 = 9.
try to make single digit by adding all the numbers when multiplied by any number
ex; 9 * 12345678 = 111111102
1+ 1+ 1+ 1+ 1+ 1+1+ 0+ 2= 9

Answer 3

Its the sum of the digits in the answer,if sum is greater than one digit you add the numbers -then add the digits in the number: such a 573x 9= 5157,add 5+1+5+7=18, add the digits of the sum 1+8=9
548976x9=4,940,784, 4+9+4+0+7+8+4=36, 3+6=9
It will always be 9.

Answer 4

For any number divisible by 9, the sum of the digits is divisible by 9.

Conversely, all numbers whose digits add up to 9 (or a multiple of 9) are divisible by 9.

And, it is not unique to 9. It works for 3 also.

It is an easy way to determine if a number is divisible by 9 (or three)

Need to know if a number is divisible by 7?
Take off the last digit and multiply it by 2. Subtract that result from the remaining digits. Is the result divisible by 7? (Repeat as necessary.)

Try 161... Take off the last digit (1), leaving 16. Multiply the 1 you removed by 2 = 2. Subtract the 2 from the remaining digits (16), to get 14.
Not sure about 14? Take off the 4, multiply it by 2 to get 8. Remaining digit is 1, subtract 8 to get -7.

Answer 5

First of all, it's not any number, it's any number between 1 and 10 (inclusive) and some others (it is true for all numbers that have 1 non-zero digit).

For example 11•9= 99, but 9+9=18≠9.

What you will find that for any number that is a multiple of 9, then 9 divides the sum of that's numbers digits.

For example 11•9=99 and 9+9=18, this is not equal to 9, but 9 divides 18.

Why does this happen?

Let (a0,a1,a2,a3, . . ., an) (where the ai's are between 0 and 9) define the number an•10^n+a(n-1)•10^(n-1)+ . . .+ a1•10+a0. The first thing you need to notice is that for any n, when you divide 10^n by 9, you have a remainder of 1. For example 10^3 = 1000 = 999+1 = 111•9+1. Therefore (10^n)-1 is a multiple of 9.

So let the number (a0,a1,a2,a3, . . ., an) be divisible by 9, then (a0,a1,a2,a3, . . ., an) = an•10^n+a(n-1)•10^(n-1)+ . . .+ a1•10+a0 is divisible by 9.
And an•10^n+a(n-1)•10^(n-1)+ . . .+ a1•10+a0 = an•(10^n-1) + a(n-1)•(10^(n-1)-1) + . . .+ a1•(10-1)+a0 +an+a(n-1)+ . . . +a1 is divisible by 9. But an•(10^n-1) + a(n-1)•(10^(n-1)-1) + . . .+ a1•(10-1) is divisible by 9 (as explained above) so an+a(n-1)+ . . . +a1+a0 must be divisible by 9. Therefore if a number is divisible by 9, then the sum of it's digits is also divisible by 9.

Answer 6

Eulercrosser is right, as was his proof. If you rephrase it as "the number 9 when multiplied by any number and allo of the digits are added together leave us with something DIVISIBLE by 9", then the statement is true. In fact, if you take any number, sum up its digits, and then subtract the original number, what you have will be divisible by 9:

Example:

n=347, sum of digits: 14, and 9 divides 347-14=333.

This process is sometimes called "casting out 9's".

There are other cool tests for divisibility by certain numbers:
1. A number is divisible by 2^n for some n only if the number
represented by its last n digits is divisible by n, i.e., 429856736
is divisible by 2^2=4 since the number 36 is divisible by 4.

2. A number is divisible by 3 only if the sum of its digits is.

3. A number is divisible by 11 only if, when you start at the last digit and alternately subtract and add the previous digits, what you have is divisible by 11, i.e., for 40249, form the sum 9-4+2-0+4=11, so 40249 is divisible by 11.

4. A number is divisible by 99 only if when you form two digit numbers from pairs of digits of your given number, starting at the end, and add those numbers together, what you get is divisible by 99. Example: for 34155, add together the numbers 55+41+03, the sum is 99, so 34155 is divisible by 99.

5. One can test a number for divisibility by 7,11,or 13 by forming 3 digits numbers starting at the end of your given number, alternately adding and subtracting these 3 digit numbers, and seeing if the result is divisible by 7,11,or 13. Example: 34424481. Form the sum 481-424+34=91. This is divisible by 7 and 13, but not 11, so the original number was divisible by 7 and 13, but not 11.

All of these tests are designed to work if your number is given in base 10. If you write your numbers in a different base (like in binary), you will have different tests. I have even heard that if you write the binary representation for pi, there is a formula that gives its nth digit. Funny things can happen if you change your base.

Answer 7

Hmmm, let's see:

First if the number has 2 digits to begin with it is not true (60 doesn't really qualify as a 2 digit number here.)

For instance 21*9 = 189 sum is 18 divisible by 9 but not 9.

Now : Suppose the number divisible by 9 is of the form ab = 9k, that is it has a value of 10a+b, so 9a+(a+b) =9k , k is an integer so a+b must also be divisible by 9.

Answer 8

A positive whole number is divisible by 9 if and only if the sum of its digits is divisible by 9. This can be proved using modular arithmetic. For the case of a 3 digit number,

ABC = A*(10)^2 + B*(10) + C,

the idea is that 10 is congruent to 1 mod 9, so

ABC is congruent to A*(1)^2+B*(1) + C=A+B+C.

In other words, ABC - (A+B+C) is a multiple of 9. If we know that ABC is a multiple of 9, it follows that A + B + C is a multiple of 9. Conversely, if we know that A+B+C is a multiple of 9, it follows that ABC is a multiple of 9.

Answer 9

Sometimes 9 is called as magic number. Anything you do with 9 and the result will always give you 9 when the digits are added.

Answer 10

Because in decimal system 9 is the last single digit number.