There are 6 major political parties in a country. An organization conducted a survey for these 6 major political parties to see what people are going to vote for. The results of the survey, in percentage terms, are as follows:

Parties Percentage (%)
PP1 13.5
PP2 15.2
PP3 35.7
PP4 20.4
PP5 8.1
PP6 4.3
Not Sure 2.8

In their calculation, they rounded percentages up or down to nearest tenth (1 decimal place) as seen above and out of blue you start to wonder at least how many people they may survey to get these percentages.

  1. What is the least number of people they survey to have these rounded percentages?


  1. Choose your own 1 decimal percentages (as shown above) to maximize the least number of people a organization to survey?

P.S.: A real life question :)

  • $\begingroup$ This looks more like a math problem than a math puzzle. I am very bad at demarcating the two. Question: If off-topic, should I delete my response? $\endgroup$
    – Cardinal
    Dec 10, 2020 at 20:29
  • $\begingroup$ I have asked a similar question some time ago. $\endgroup$
    – WhatsUp
    Dec 11, 2020 at 8:35

3 Answers 3


This feels like a math problem more than a math puzzle.

Part 1


Part 2

(0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 98.2) for 2001

(1, 3, 5, 7, 9, 11, 1965)

Part 1 by brute force


for individual groups of 73, 82, 193, 110, 44, 23, 15


Using a spreadsheet, make the lower bound = the rounded percent - 0.05. The upper bound = the rounded percent + 0.0499999999. The lower bound x N and the upper bound X N must have an integer between them.

To save myself time of checking each number, I used the floor and ceiling function in Excel. The floor of the upper bound minus the ceiling of the lower bound must equal zero or there is not an integer between them. Intuitively, 1000 was a hard cap because multiplying each percent by 10 will give all integer values. An if statement could then be used to make sure all 7 groups (6 parties and not sure) were true. 540 was the lowest number meeting this criteria.

Part 2: Credit to Bubbler's answer for providing the logic. I thought I had a better number, but the rounding could be simplified.

As Bubbler points out, the lowest denominator which rounds down to 0 is 1/2001. The -0.3% under 100% for a sum looks like a limit. Finally, this officially turns into a math problem and not a puzzle. There are many possible combinations that give 2001.


Answer to part 2:



If all cases have either 0 an odd number of the 2000 votes, and not all numbers are multiples of 5.


As long as all numbers are rounded up, we can deduce the exact percentages from the fact that the numbers add up to more than 100%

Simple example:

Percentages: 100%; 0.1%; 0%; 0%; 0%; 0%; 0%;
100% must be rounded up from 99.95% and 0.1% rounded up from 0.05% to make the sum of 100.1% possible, and 0.05% must be at least 1 vote. This makes the number of people a multiple of 2000.


Part 2 answer:

2001, using the percentages 99.7%, 0%, 0%, 0%, 0%, 0%, 0%. (The highest voted item doesn't matter as long as it's between 99.7% and 99.9% inclusive.)


Given that the total of the values is too low, some fractions must be distributed to the zeros to make the total of 100%. The fraction with the lowest denominator which rounds down to 0 is 1/2001 = 0.0004997... < 0.0005.

  • $\begingroup$ Hmm, I get a minimum of 2000, 2400, and 4000 for 99.7%, 99.8%, and 99.9%, respectively. $\endgroup$
    – RobPratt
    Dec 11, 2020 at 14:54
  • $\begingroup$ @RobPratt I think this is right (assuming everyone falls in one of the categories, and rounding down .5 is disallowed). 99.9 would be e.g.1999,1,1,0,0,0,0 votes $\endgroup$
    – Retudin
    Dec 11, 2020 at 17:15
  • $\begingroup$ Yes, I guess the discrepancy occurs because of the ambiguity of whether 0.5 is rounded up or down. $\endgroup$
    – RobPratt
    Dec 11, 2020 at 18:45
  • $\begingroup$ @retudin I agree. The distribution I put forward could be simplified down to this sample size after re-evaluating and that's if I didn't miss something else. The range appears to extend beyond 99.7-99.9 unless I made another error. $\endgroup$
    – Cardinal
    Dec 11, 2020 at 19:45

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