More Math for Fun

[Brickjectivity]

Oops I misread the directions.
My new answer:

Spoiler: New & Improved

1 pushup clap, and then 3599 seconds remain. If he pushup-claps every 60 seconds then it is:
1 + (3599 - 3599 mod 60)/60 = 60.

 

[Debater Slayer]

Spoiler: workout

Initially one clap and one pushup, and then every 15 seconds a pushup with a clap every 20 seconds. His workout ends in exactly 3600 seconds, so the last clap and pushup within that time is what we will count up to. We can assume that the first clap and pushup take part of the first second or a whole second, which leaves 3599 seconds for the rest of the workout.


Pushups: 1 + (3599 - 3599 mod 15)/15 = 240
Claps: 1 + (3599 - 3599 mod 20)/20 = 180

I tried to find a name for the above operation of f() = (M - M mod N)/N. I couldn't find a name for it. It means "Count as many as will fit and round off."

When X = M/N such that X contains decimals, rounding it down to the nearest integer is called floor division. Rounding it up is called ceiling division.

 

[Debater Slayer]

The answer is 61! Most of the answers either got it right or got very close.

The LCM of 15 and 20 is 60. This means the push-up and clapping coincide every 60 seconds. In an hour, this works out to 3600/60 = 60 times. However, Joe also did both simultaneously at the very start of the workout, so the answer is 61.

If I say that I'll perform an action every three seconds for 12 seconds, this gives one at the start, one at t = 3 seconds, one at t = 6 seconds, one at t = 9 seconds, and then the final one at t = 12 seconds, for a total of five actions. So the number of actions is t/r + 1, where t is the total time and r is the duration of the interval between each action.

 

[Subduction Zone]

The answer is 61! Most of the answers either got it right or got very close.

The LCM of 15 and 20 is 60. This means the push-up and clapping coincide every 60 seconds. In an hour, this works out to 3600/60 = 60 times. However, Joe also did both simultaneously at the very start of the workout, so the answer is 61.

If I say that I'll perform an action every three seconds for 12 seconds, this gives one at the start, one at t = 3 seconds, one at t = 6 seconds, one at t = 9 seconds, and then the final one at t = 12 seconds, for a total of five actions. So the number of actions is t/r + 1, where t is the total time and r is the duration of the interval between each action.

That was the clarity that I wanted. The wording implied to me that by the end of each 15 second period he would have done a pushup. My Joe was just a little bit on the lazy side.


EDIT: I reread the OP and dang it, he does one right at the start. My Joe owes you one clapping push up.