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.....

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- Nov 29th 2010, 06:08 PMjpmath2010ABC 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, 07:50 PMSoroban
Hello, jpmath2010!

Is that the exact wording of the problem?

As stated. it's a silly problem . . .

Quote:

$\displaystyle \text{Let }ABC\text{ be a 3-digit number such that its digits }A, B, C$

. . . $\displaystyle \text{form an arithmetic sequence.}$

$\displaystyle \text{The largest integer that divides all numbers of form }ABC\!ABC\text{ is ... ?}$

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

- Nov 29th 2010, 09:26 PMelemental
$\displaystyle 3003$.

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, 03:35 AMjpmath2010
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, 08:52 AMSoroban
Hello, jpmath2010!

Now I get it . . .

Quote:

Let $\displaystyle ABC$ be a 3-digit number such that its digits form and arithmetic sequence.

Find the largest integer that divides all numbers of form $\displaystyle ABC\!ABC$

Let $\displaystyle \,h$ = hundreds digit of the three-digit number,

. . and $\displaystyle \,d$ be the common difference of the arithmetic sequence.

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

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

Therefore, the largest divisor is $\displaystyle 3003.$

- Nov 30th 2010, 04:08 PMjpmath2010
wow thats great....i really thank you so much....have a nice day and God bless...