Why are these numbers always evenly divisible by 13?

Question

Take any 3-digit number, and repeat it to form a 6-digit number. For example, choose 549 and repeat it to form 549549. Why are these 6-digit numbers always evenly divisible by 13?

Answer 1

Your number is of the form:
abcabc

Written algebraically, you have.
1000 * 'abc' + 'abc'
1000 'abc' + 1 'abc'
(1001) 'abc'

In other words, the combined number 'abcabc' is 1001 times the value of 'abc'.

If you check, you'll find that 1001 is evenly divisible by 13 (and 7 and 11).

Answer 2

You can split up any number into multiples of 10. For example, 549 = 500 + 40 + 9.

So a six digit number like that is going to have the value of

a*10^5 + b*10^4 + c*10^3 + a*10^2 + b*10 + c

Combining like terms, you get

a(10^5 + 10^2) + b*(10^4 + 10) + c*(10^3 + 1)

Factor out some powers of 10 and you get:

a(10^2)(10^3 + 1) + b(10)(10^3 + 1) + c*(10^3 + 1) =
(10^3 + 1) * (a(10^2) + b(10) + c)

So the number is always a multiple of (10^3 + 1), or 1001. 1001 is a multiple of 13, therefore the entire number is always a multiple of 13, regardless of the values of a,b, and c.

Answer 3

Any 3-digit number XYZ can be written as A*13 + B (where B is 0 when it is a multiple of 13).
Now, if you make a new 6-digit number as you've described, then this number is actually
(A*13+B)*1000 + A*13+B
because you're basically multiplying XYZ by 1000 and adding to itself to the product.
Simplifying, the above can be written as
A*(13000+13) + B*(1000 + 1)
or
A*130013 + B*1001
Now, 130013 is a multiple of 13 (it is 1001*13)
And 1001 is a multiple of 13; it is (13*77)
Therefore, we can write the above as
(A*1001 + B*77)*13, which is a multiple of 13.

Since we didn't make any assumptions about A and B, this should apply to all 3-digit numbers.

Answer 4

Let the 3 digit number be abc.
Repeated yo uget abcabc which equals abc=1001.
1001=7*11*13 so any number of the form abcabc is divisible by 7, 11 & 13.

Also not that for a number abcdef, if abs(abc-def) is divisible by 7, 11 or 13, so is the original nunber.

Answer 5

The number is of the form 1000X+X which is equal to 1001X. Since 1001 is divisible by 13 the formula is true for any X.

Answer 6

The number can be written as:
100000h+10000t+1000u+100h +10t + u
= 100,100h + 10010t + 1001u
= 13( 7700h + 770t + 77u)
So clearly the number is divisible by 13 regardless of the values of h,t,and u.

Answer 7

The number would be 100a+10b+c(a,b,c are the digits)

If you write the6 digit number it would be

100,000a +10,000b +1000c +100a +10b+c= 100,100a +10,010b+ 1001c = 13 ( 7700a+770b +77c) which for any digits a,b,c is a multiple of 13

Answer 8

Because you are actually multiply the number by 1,001 and that is equal to 77 x 13.

Answer 9

by 1!!!!