4

$$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$

Find x

My younger sister in grade 5 had this question in a test. But I, a college student, still can't solve this. What a shame :<

Jeel Shah
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LamNS
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    So your sister is 10 to 11 years old (grade 5). I am curious as to what school is asking this question to 10 year olds. Did your little 10 year old sister give you any hints? – J. W. Perry Dec 05 '13 at 08:50
  • No hint for me.

    This ques is in an entrance exam practice test of the best secondary school in my country. At first I just gave it a try, but it's much more harder than I think. I just got 3/10 on that test.

     – LamNS Dec 05 '13 at 09:17

3 Answers3

2

One can group the $100$ summands on the right side by the ending digit of the denominators. For example, taking the numbers ending in 2:

$$\frac{1}{2\times 12}+\frac{1}{12 \times 22}+ ... +\frac{1}{92 \times 102}=\\ = \frac{1}{10}\left(\frac{1}{2} -\frac{1}{12}\right) +\frac{1}{10}\left(\frac{1}{12} -\frac{1}{22}\right) +\cdots +\frac{1}{10}\left(\frac{1}{92}- \frac{1}{ 102}\right) =\\ =\frac{1}{10}\left(\frac{1}{2}-\frac{1}{102}\right)=\\ =10\left(\frac{1}{2 \times 102}\right)\\ $$

Therefore, $x=10$

This can be formalized, of course. But I don't think that this (even without the formalization) is appropiate for a 10 years old...

leonbloy
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1

Purely computationally:

$$\sum_1^{10} \frac{1}{k(k+100)} = \frac{59810902182173}{2110393648079145}, $$ and $$\sum_1^{100} \frac{1}{k(k+10)} = \frac{119621804364346}{422078729615829}. $$

The second numerator is exactly double the first, and the second denominator is exactly a fifth of the first, leading to the surprising (for me) result that $$x=10.$$

Peter Phipps
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1

$$\frac{1}{k(k+10)}=\frac{a}{k}+\frac{b}{k+10}$$

$$\begin{array}{l} \frac{a}{k}+\frac{b}{k+10}=\frac{k(a+b)+10a}{k(k+10)} \end{array}$$

$$k\leftarrow 0, a=\frac{1}{10}$$

$$k\leftarrow 1, b=-\frac{1}{10}$$

$$\boxed{\cfrac{1}{k(k+10)}=\cfrac{1}{10}\left(\cfrac{1}{k}-\cfrac{1}{k+10}\right)}$$

$$\begin{array}{l} \sum_{k=1}^{100} \frac{1}{k(k+10)}&=\frac{1}{10}\sum_{k=1}^{100} \frac{1}{k}-\cfrac{1}{k+10}\\ &=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=1}^{100}\cfrac{1}{k+10}\right)\\ &=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=11}^{110}\cfrac{1}{k}\right)\\ &=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=101}^{110}\cfrac{1}{k}\right)\\ &=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=1}^{10}\cfrac{1}{k+100}\right)\\ &=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\cfrac{1}{k+100}\right)\\ &=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{100}{k(k+100)}\right)\\ \sum_{k=1}^{100} \frac{1}{k(k+10)}&=10\sum_{k=1}^{10} \frac{1}{k(k+100)} \end{array}$$

xavierm02
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