Machine releases a ball bearing every 7056 seconds and another machine releases a ball bearing every 1666780 seconds if they drop a ball bearing together on March 1st at 8 p.m. then the next time they will drop a ball bearing at the same time is that 12 minutes after midnight on
So what is the answer?
its may 17th
To find out when the two machines will drop a ball bearing together again, we need to determine the least common multiple (LCM) of the two time intervals.
Machine 1 releases a ball bearing every 7056 seconds.
Machine 2 releases a ball bearing every 1666780 seconds.
To find the LCM, we can use the prime factorization method:
Prime factorization of 7056:
7056 = 2^4 * 3^2 * 7^2
Prime factorization of 1666780:
1666780 = 2^2 * 5^2 * 59 * 563
Now, we take the highest power of each prime factor from both factorizations:
2^4 * 3^2 * 5^2 * 7^2 * 59 * 563
Calculating the LCM, we multiply all the prime factors together:
LCM = 2^4 * 3^2 * 5^2 * 7^2 * 59 * 563
LCM ≈ 33592876860 seconds.
Next, we need to calculate the time from March 1st at 8 p.m. until 12 minutes after midnight, which is 4 hours and 12 minutes.
4 hours = 4 * 60 * 60 = 14400 seconds
12 minutes = 12 * 60 = 720 seconds
Total time = 14400 seconds + 720 seconds = 15120 seconds.
Finally, we add this total time to the starting time of March 1st at 8 p.m.
Calculating the time when the two machines will drop a ball bearing together again:
Time = March 1st at 8 p.m. + 15120 seconds + LCM
Therefore, the next time the two machines will drop a ball bearing together is approximately 12 minutes after midnight on March 2nd.
I don't understand
their LCM is 2,940,199,920
2940199920s * 1day/86400s = 34030 days 11 hr 12 min
Maybe that will help answer your question
hint: look at the possible answers on the sheet
C'mon, man. 34030 days = 93 yrs 85 days.
So count 'em off from March 1st. Adjust for leap years if you want.
And it won't be 12 min after midnight, since it is 11 hrs after 8 pm.