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.....
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.....
Hello, jpmath2010!
Is that the exact wording of the problem?
As stated. it's a silly problem . . .
$\displaystyle \text{Let }ABC\text{ be a 3-digit number such that its digits }A, B, C$
. . . $\displaystyle \text{form an arithmetic sequence.}$
$\displaystyle \text{The largest integer that divides all numbers of form }ABC\!ABC\text{ is ... ?}$
The largest integer that divides $\displaystyle ABC\!ABC$ is . . . $\displaystyle ABC\!ABC.$
Hello, jpmath2010!
Now I get it . . .
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 $\displaystyle 3003.$