When Camera A and Camera B take their last picture together, how many minutes before noon is it?

Question

There are two cameras that take pictures of a traffic intersection. Camera A starts taking pictures at $6$ AM and takes a picture every $11$ minutes. Camera B starts taking pictures at $7$ AM and takes pictures every $7$ minutes. Camera A and Camera B take a picture at the same time at four different times before noon. When Camera A and Camera B take their last picture together, how many minutes before noon is it?

Hi! I am a student (currently middle school), so simple explanations would be very appreciated. I tried writing a modular congruence to solve this problem.

First, I noticed that there is a 6 minute difference between the two times if we start at 7:00. For example, when Camera B starts taking pictures at 7:00, Camera A will take a picture at 7:06. Therefore, the times that they take pictures at the same time can be written as an equation: $7x = 11y + 6$.

Written as a congruence, I think this means that $7x \equiv 6 \pmod{11}$. Then, I multiplied both sides by 8 to get the following expression: $56x \equiv 48 \pmod {11}$. This can be reduced to $x \equiv 4 \pmod {11}$.

What steps should I take now? Was my initial approach correct?

Answer 1

Your initial approach is great:

$x = 4\pmod{11}$

The first time the cameras sync up is at 7:28

And since $\gcd(7,11) = 1$ the the cameras will sync up every 77 minutes.

Noon is $5\cdot 60 - 28 = 272$ minutes after the first time the cameras sync

$\lfloor\frac {272}{77}\rfloor = 3$

The cameras will synch up 3 more times before noon.

at 8:45, 10:02, 11:19

Answer 2

Yes, you've done great so far!

From what you've already written (the cameras sync up exactly when $x \equiv 4 \mod 11$) we can say that when $x= 4, 15, 26, 37$ the cameras take a picture together.

When $x=4$, the time is $7$ AM + ($4*7$) minutes = $7:28$ AM

When $x = 37$, the fourth and final time that the cameras sync up, the time is $7$ AM + ($37*7$) minutes = $11:19$ AM