Answer:
true
Step-by-step explanation:
there is a nice rule for when a number is divisible by 13 :
starting with the lowest digit value (the last on the right) form blocks of 3 digits as numbers and give them alternating signs (starting with +).
if the last block (the last on the left) has less than 3 digits, just use the remaining ones as a number.
add all these numbers (with their associated signs).
if the number calculated by this (alternative) sum is divisible by 13 then the original number is divisible by 13.
if N = abcdefghi, then (ghi - def + abc) must be divisible by 13, so that N is also divisible by 13.
in our case, N = abcabc.
the block calculation is therefore abc - abc = 0, which is always divisible by 13.
so, any number formed like this must be divisible by 13.
FYI - a quick look at why that is :
N = abcabc =
= a×10⁵ + b×10⁴ + c×10³ + a×10² + b×10 + c =
= a×(10⁵ + 10²) + b×(10⁴ + 10) + c (10³ + 1) =
= a×100100 + b×10010 + c×1001
define d = 1001 = 77×13
and we get
a×d×100 + b×d×10 + c×d
a sum is divisible by a number, if every summed up term is divisible by that number.
since d is a factor in every term, and d is divisible by 13, so is every term and therefore the whole number.
so, every number in the form of abcabc is divisible by 13.
