Thread: ABC is a 3-digit number such that its digits A, B and C form an arithmetic sequence..

1. ABC is a 3-digit number such that its digits A, B and C form an arithmetic sequence..

Let ABC be a 3-digit number such that its digits A, B, and C form and arithmetic sequence. The largest integer that divides all numbers of form ABCABC is.....?

please give me the complete solution of this problem.....

2. Hello, jpmath2010!

Is that the exact wording of the problem?
As stated. it's a silly problem . . .

$\text{Let }ABC\text{ be a 3-digit number such that its digits }A, B, C$
. . . $\text{form an arithmetic sequence.}$
$\text{The largest integer that divides all numbers of form }ABC\!ABC\text{ is ... ?}$

The largest integer that divides $ABC\!ABC$ is . . . $ABC\!ABC.$

3. $3003$.

Soroban, he is asking what number divides ALL numbers of this form. For example,
the GCD of 123123 and 135135 is 3003. The GCD of 147147 and 963963 is also 3003.

4. thank you soroban for your effort....Any way the problem i have posted is come the Mathematics Olympiad.That is the exact problem. would you please give me the exact solution of that problem.

5. Hello, jpmath2010!

Now I get it . . .

Let $ABC$ be a 3-digit number such that its digits form and arithmetic sequence.
Find the largest integer that divides all numbers of form $ABC\!ABC$

Let $\,h$ = hundreds digit of the three-digit number,
. . and $\,d$ be the common difference of the arithmetic sequence.

Then: . $ABC \;=\;100h + 10(h+d) + h+2d \:=\:111h + 3d \:=\:3(37h + d)$

Hence: . $ABC\!ABC \;=\;1001\cdot ABC \;=\;1001\cdot 3(37h + d) \;=\;3003(37h + d)$

Therefore, the largest divisor is $3003.$

6. wow thats great....i really thank you so much....have a nice day and God bless...