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In a certain medical group, Dr. Schwartz schedules appointments to begin 30 minutes apart, Dr. Ramirez schedules appointments to begin 25 minutes apart, and Dr. Wu schedules appointments to begin 50 minutes apart. All three doctors schedule their first appointments to begin at 8:00 in the morning, which are followed by their successive appointments throughout the day without breaks. Other than at 8:00 in the morning, at what times before 1:30 in the afternoon do all three doctors schedule their appointments to begin at the same time?

A. 9:30 in the morning

B. 10:00 in the morning

C. 10:30 in the morning

D. 11:00 in the morning

E. 11:30 in the morning

F. 12:00 noon

G. 12:30 in the afternoon

H. 1:00 in the afternoon

I've solved the problem before by drawing out clocks or writing out the progression of time in tabular form.

This problem seems related to modular arithmetic or algebra.

I'm looking for some modular arithmetic explanation for this as an alternative way of thinking about it or some drawings of clocks spinning to confirm my approach to solve this problem. I don't trust clock drawings so I'd rather request the algebraic way of doing this.

This isn't a duplicate because I specifically will reject anything relating to LCM and isn't related to "mod" or diagrams.

Community
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  • a lot of modular arithmetic involves gcd and lcm. –  Aug 13 '19 at 13:12
  • Not clear why you wish to reject lcm properties (which were designed exactly for such). In any case I added an answer with and without for comparison. – Bill Dubuque Aug 13 '19 at 14:05

3 Answers3

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The problem is equivalent to determining the smallest positive integer $x$ (# of minutes) such that

$$x \equiv 0 \pmod{30} \tag{1}\label{eq1}$$ $$x \equiv 0 \pmod{25} \tag{1}\label{eq2}$$ $$x \equiv 0 \pmod{50} \tag{1}\label{eq3}$$

Using the LCM (I assume that's what you mean by LCD) is the way to normally do this, but as you don't want to use that, here's an alternate method. Note that at every $50$ minutes, the appointments by Dr. Ramirez and Dr. Wu occur at the same time, but not at any other times. Thus, you want to find where a multiple of $50$ and $30$ occur. By checking, you see that $50$ and $100$ don't work, but $150$ does, so it'll take $150$ minutes $= 2$ hours, $30$ minutes. This is answer (C), i.e., $10$:$30$ a.m. for the first time and, as Matthew Daly's comment pointed out, every multiple of $150$ minutes works too, so there's also a next time of $1$:$00$ p.m. (i.e., answer (H)) before $1$:$30$ p.m.

John Omielan
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Let $x$ be a number of minutes after 8:00 at which all three doctors have appointments scheduled. Then $$x\equiv 0\pmod{30}$$ $$x\equiv 0\pmod{25}$$ $$x\equiv 0\pmod{50}$$ You can solve this system of congruences with the Chinese Remainder Theorem to see that the solution is $x\equiv 0\pmod{150}$.

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Without any knowledge of LCM properties:

$30,25,50\mid n\iff 30,50\mid n\iff 10\mid n\ \ \&\ \ \smash{\underbrace{3,5\mid n/10}_{\large \llap{\iff} 15\ \mid\ n/10 }}\iff 150\mid n$

Using basic LCM laws:

$[30,25,50] = [30,50] = 10[3,5] = 10[15] = 150$

Bill Dubuque
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