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I am reading through this proof and I have several questions-

  1. In the 4th paragraph, it says $111 \times 5 = 555$ why is $111$ being multiplied by $5$? Same goes for why $55$ is multiplied by $5$?

  2. In the same paragraph, $55$ is subtracted from $555$ and then $1$ is added- why does $1$ need to be added?

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I know this question has been asked before on stack exchange and I have looked at the link A question related to Pigeonhole Principle. But I am hoping to understand this particular approach.

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user2553807
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  • The maximum sum their ages 60 + 59 + 58 +... =555. In the days before calculators if you were given a sum of 10 terms and just told that it just happens to equal 555, it'd be expected the reader would ask how you calculates that. Well, 60 + 51 =111 and 59 + 52 = 111 and... that's how they added all the numbers up to get 555. Second question. How many numbers are there between 555 and 55 including both the 55 and 555. 555 -55 is the number not including 55. So 555-55+1 is the number including 55. Try it with how many numbers between 6 and 10. There are 5 (6,7,8,9,10). 5 =10-6+1. – fleablood Feb 20 '16 at 22:46

1 Answers1

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For your first question, the author is rearranging the sum $$60+59+58+57+56+55+54+54+52+51$$ as

$$(60+51)+(59+52)+(58+53)+(57+54)+(56+55)\;;$$

each of the single sums in parentheses is $111$, and there are $5$ such sum, so the total is $111\cdot5$. The author is doing the same thing with the other sum:

$$(1+10)+(2+9)+(3+8)+(4+7)+(5+6)=11\cdot5\;.$$

For your second question, notice that there are $555$ integers in the range from $1$ through $555$ and $55$ in the range from $1$ through $55$. Thus, there are $555-55$ in the range from $\mathbf{56}$ through $555$: by subtracting $55$ we’ve thrown away $1,2,3,\ldots,55$. But we didn’t want to throw away $55$: we wanted the number of integers in the range from $\mathbf{55}$ through $555$, so we have to add $1$ to count the $55$.

Alternatively, you can argue that from the set $\{1,2,\ldots,555\}$, which has $555$ members, we need to throw away $\{1,2,\ldots,54\}$, which has $54$ members, so we’re left with

$$555-54=555-(55-1)=555-55+1$$

numbers.

Brian M. Scott
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