A bloggers' team needs to process two video files: one twice the size of the other.

In the first half of the night, the whole team worked on the large video file. Then, after midnight, they split into two equal parts (split in half):

  • the first half of team continued to work on the large video file and at morning it was done, and
  • the second half of team began working on the small video file and in the morning it remained to process the part that one blogger can do in one full night.

Question: How many bloggers were on the team?

The original story was borrowed from Perel'man's book.

  • $\begingroup$ Is "split in half" a precise statement here? Because after midnight the unprocessed size of the bigger file and the smaller one is the same, and if the same number of bloggers are working on them, they should be done at the same time. Not sure what I am missing here. $\endgroup$
    – justhalf
    Commented Jul 13, 2020 at 1:41
  • 1
    $\begingroup$ "split in half" = "split into two equal parts" is correct here. The error was in "and at midnight half of this file was ready". I have edited the post. $\endgroup$
    – Nick
    Commented Jul 13, 2020 at 2:25
  • $\begingroup$ Just to clarify - are two halves of the night split by midnight? $\endgroup$
    – Quintec
    Commented Jul 13, 2020 at 2:34
  • $\begingroup$ @Quintec, yes, you are correct. $\endgroup$
    – Nick
    Commented Jul 13, 2020 at 4:14

3 Answers 3


There were a total of

8 bloggers

We can think of the files as requiring a certain amount of man-hours to complete. The large file was completed by

All the bloggers for half the night and then half the bloggers for half the night. Let's call the total number of bloggers 2Y and lets call the total hours in the night 2T. Then the statement above boils down to this many man-hours: 2Y * T + Y * T = 3YT
Since the smaller file is half the size of the larger file it will require half the man-hours, so 3/2 * YT. We know that half the bloggers worked on it for half the night so already YT man-hours have been performed and still 1 full night with one blogger is needed (1 * 2T = 2T). So the final equations is this: 3/2 * YT = YT + 2T. This simplifies to Y = 4 and 2Y = 8 => the total number of bloggers.

  • $\begingroup$ ^vote with a note: This solution could be slightly more streamlined with Y = number of bloggers instead of Y = 1/2 number of bloggers. (Unlike T = 1/2 night hours, as is, which is at least as streamlined than would be T = night hours.) $\endgroup$
    – humn
    Commented Jul 13, 2020 at 16:05

If we think in 1/2 night shifts.

1. The large file had everyone working for 1 shift, and 1/2 working for another shift. That's 1.5 shifts (1/2 nights) for the whole team.

2. The small file had 1/2 the team working for one shift and still had 2 shifts to go for a single person.

Using 2, the number of shifts for the small file is (a/2 + 2) where a is the number of people on the team.

Using 1, the number of shifts for the large file is a + a/2 or 3a/2.

Since the smaller file is 1/2 the size, it should require 1/2 the work. Let's multiply it by 2 then and set equal to the same work done in the large file:
a + 4 = 3a/2

Subtract a from both sides gives us a/2 = 4, multiply both sides by 2 gives us a = 8, so there were 8 bloggers on the team.


Had the big video as 2V . The whole team made 2/3 of 2V in 1/2 night = 4V/3. That means 1/2 the team makes 2/3 of V in 1/2 a night.So the unfinished side have 1/3 of V left , the amount 1 blogger makes in a night. We know this whole team can make 2 x 4V/3 = 8/3 in 1 night. That means 8 guys in the team.

  • $\begingroup$ This was already correctly answered by Amorydai 18 hours ago. $\endgroup$ Commented Jul 13, 2020 at 21:42
  • $\begingroup$ @Randal'Thor, I thought the spoilers are there so that people could post their answer without reading answers of others. Is this not what spoilers are for? $\endgroup$ Commented Jul 14, 2020 at 3:04
  • $\begingroup$ I'm new - I presumed that multiple answers were acceptable as we see a variety of ways of problem solving $\endgroup$
    – DrTris
    Commented Jul 30, 2020 at 13:25

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