I have the following question:

Let $N$ be the greatest number that will divide $1305, 4665$ and $6905$, leaving the same remainder in each case. What is the sum of digits of $N$.

My approach was to first of all find the LCM of these $3$ numbers which is say $x$ and add $9$ to it.

This approach is absolutely incorrect.

  • 3
    $\begingroup$ I do not see a clear path that starts that way. There are some natural equations one can write down, and that lead quickly to the answer. $\endgroup$ – André Nicolas Dec 14 '13 at 18:16
  • $\begingroup$ oh sorry my approach is just absolutely wrong. $\endgroup$ – iajnr Dec 14 '13 at 18:18
  • $\begingroup$ Oops. Saw @Andre 's comment only after posting. Apologies, I believe you were trying to guide using hints and I messed that up. $\endgroup$ – Aryabhata Dec 14 '13 at 18:38
  • $\begingroup$ @Aryabhata: No problem, answers were bound to come quickly. Anyway, the hint may have done its job. $\endgroup$ – André Nicolas Dec 14 '13 at 18:41

If the three numbers are $a,b,c$, I believe the answer you are looking for would be $\text{gcd}(a-b, b-c)$.

Say numbers are $a_1, a_2, a_3$. Then we have that for some $x$

$$a_i = q_i N + x$$

This implies that $N$ dividies $a_2 - a_1$ and $a_3 - a_2$, and any $k$ which divides both the differences will necessarily leave the same remainder with the three numbers.

Thus you take the greatest common divisor for the two differences.


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