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A novel has 6 chapters. As usual, starting from the first chapter begins on a new page. The last chapter is the longest and the page numbers of its pages add up to 2010: How many pages are there in the first 5 chapter?

Let the pages be numbered

$$1,\dots,X_{1}; X_{1}+1,\dots,X_{2}; X_{2}+1,\dots,X_{3}; X_{3}+1,\dots,X_{4}; X_{4}+1,\dots,X_{5}; X_{5}+1,\dots,X_{6}$$

Sum of the last chapter pages $= [(X_{6}+X_{5}+1)(X_{6}-X_{5})]/2 = 2010$

(Note sum of arithmetic series with $a_{1} = X_{5}+1$ and $a_{n} = X_{6}$)

Now $2\cdot 2010 = 4020 = 67\cdot 60 = 134\cdot 30 = 268\cdot 15$

Thus $X_{6}+X_{5}+1 = 67$ and $X_{6}-X_{5} =60$, which gives $X_{6} = 63$, and $X_{5} = 3$. This is not possible as there are five chapters.

Next $X_{6}+X_{5}+1 = 134$ and $X_{6}-X_{5} = 30$ gives non-integer. So this is not possible.

Next $X_{6}+X_{5}+1 = 268$ and $X_{6}-X_{5} = 15$ gives $X_{6} = 141$, and $X_{5} = 126$. This seems plausible but it does not pass the test that chapter 6 is the longest as you could have a longer chapter when $X_{5} = 126$. Any other product of factors would still not hold the condition of the "Longest Chapter". What seems to to be answer for this?

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I tried MathJax syntax, Seems like it has not worked, Could someone help me with subscripts –  satish ramanathan Nov 11 '13 at 19:07
1  
The dollar sign \$ is the magic character to enter math mode ;) –  filmor Nov 11 '13 at 19:12
    
More accurately: enclose math signs (or a line of them) between dollar signs, and if you want a separate line, use double dollar signs. See the difference between $\;x_1=12x_2\;$ and $$x_1=12x_2$$ –  DonAntonio Nov 11 '13 at 19:16
    
Thanks Filmor. I will get to start using it in the future. –  satish ramanathan Nov 11 '13 at 19:18
    
Note that "you could have a longer chapter" is not the right approach here. The last situation is actually impossible, as the first $5$ chapters have on average $25.2$ pages, which means one must have more than $25$ pages, which means it is impossible for the last to be the longest.... –  N. S. Nov 11 '13 at 19:31

1 Answer 1

up vote 2 down vote accepted

You missed some factorizations. For example

$$4020=201*20$$

$$X_{6}+X_{5}+1 = 201 \,;\, X_6-X_5=20 \Rightarrow X_6=110, X_5=90$$

And I think, there are few more you missed.

Note

$$4020= 2^2*3*5*67$$ has $(2+1)(1+1)(1+1)(11)=24$ divisors, thus there are exactly $24/2$ ways of writing it as a product of two integers. Anyhow, since the two factors must have different parity, only $(1+1)(1+1)(1+1)=8$ ways [namely one factor has to be $4$ times a divisor of $3*5*67$] are relevant for this problem.

Note To solve the problem, you need to check all the $8$ cases. Note that in a case, if you get $ X_6-X_5 < \frac{X_5}{5}$ this tells you that it is impossible for the last chapter to be the longest.

You are hoping to get your answer by elimination. If you can prove that 7 cases are impossible, since the problem is telling you there is one possibility by elimination it has to be the remaining case...

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Still, Does this combination ensure longest chapter? –  satish ramanathan Nov 11 '13 at 19:27
    
@satishramanathan The last chapter has 20 pages. The first 5 chapters have an average of $18$ pages, so in this case it is possible... Note that "ensures" is not the right choice of words, this answer is possible, but not guaranteed. The problem tells us that one situation is possible, if this is the only one it MUST be the case.... –  N. S. Nov 11 '13 at 19:29
    
You are right, for any combination below it, it simply cannot hold and this seems to the only one. Thanks. I really do not know how I missed this combination. Anyway, it is good to have a clarification –  satish ramanathan Nov 11 '13 at 19:38

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