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

• Nov 29th 2010, 07:08 PM
jpmath2010
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.....
• Nov 29th 2010, 08:50 PM
Soroban
Hello, jpmath2010!

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

Quote:

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

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

• Nov 29th 2010, 10:26 PM
elemental
$3003$.

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.
• Nov 30th 2010, 04:35 AM
jpmath2010
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.
• Nov 30th 2010, 09:52 AM
Soroban
Hello, jpmath2010!

Now I get it . . .

Quote:

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 $3003.$

• Nov 30th 2010, 05:08 PM
jpmath2010
wow thats great....i really thank you so much....have a nice day and God bless...