MHB Solving Sequence Problem with Elementary Math Concepts

[anemone]

Hi MHB,

My nephew, age 9, was asked the following question and he hoped I could solve the problem and then explain the solution to him using only elementary math concepts. My boyfriend has solved it, but he used a formula that he recalled seeing in a textbook by G.H. Hardy, and that involved knowledge of triangular numbers.

So, I don't know how to attack this problem from a 9 year old's perspective. If you happen to know how to simplify the problem, could you please enlighten me? Many thanks in advance.

Problem:
In the sequence of fractions $\dfrac{1}{1},\,\dfrac{2}{1},\,\dfrac{1}{2},\, \dfrac{3}{1},\,\dfrac{2}{2},\,\dfrac{1}{3},\, \dfrac{4}{1},\,\dfrac{3}{2},\,\dfrac{2}{3},\, \dfrac{1}{4},\, \dfrac{5}{1},\, \dfrac{4}{2},\, \dfrac{3}{3},\, \dfrac{2}{4},\,\dfrac{1}{5}\cdots$, fractions equivalent to any given fraction occur many times. For example, fractions equivalent to $\dfrac{1}{2}$ occur for the first two times in positions 3 and 14. In what position is the fifth occurrence of a fraction equivalent to $\dfrac{3}{7}$?

 

[mente oscura]

Hi MHB,

My nephew, age 9, was asked the following question and he hoped I could solve the problem and then explain the solution to him using only elementary math concepts. My boyfriend has solved it, but he used a formula that he recalled seeing in a textbook by G.H. Hardy, and that involved knowledge of triangular numbers.

So, I don't know how to attack this problem from a 9 year old's perspective. If you happen to know how to simplify the problem, could you please enlighten me? Many thanks in advance.

Problem:
In the sequence of fractions $\dfrac{1}{1},\,\dfrac{2}{1},\,\dfrac{1}{2},\, \dfrac{3}{1},\,\dfrac{2}{2},\,\dfrac{1}{3},\, \dfrac{4}{1},\,\dfrac{3}{2},\,\dfrac{2}{3},\, \dfrac{1}{4},\, \dfrac{5}{1},\, \dfrac{4}{2},\, \dfrac{3}{3},\, \dfrac{2}{4},\,\dfrac{1}{5}\cdots$, fractions equivalent to any given fraction occur many times. For example, fractions equivalent to $\dfrac{1}{2}$ occur for the first two times in positions 3 and 14. In what position is the fifth occurrence of a fraction equivalent to $\dfrac{3}{7}$?

Hello.(Whew)(Whew)

Let \ \dfrac{A_n}{B_m}

Let \ n=order \ number \ suscession \ numerator

Let \ x=number \ numerator

And:

Let \ m=order \ number \ suscession \ denominator

Let \ y=number \ denominator

Then:

x \ number \ position=\dfrac{n^2+(2x-1)n+x^2-3x+2}{2}. (1)

y \ number \ position=\dfrac{m^2+(2y-3)m+y^2-y+2}{2}. (2)

Making: x=m \ and \ y=n \rightarrow{}x \ number \ position=y \ number \ position

Example:

\dfrac{3}{7}=\dfrac{15}{35}

therefore:

x=15 \ and \ y=35

For (1): x \ number \ position=\dfrac{35^2+(2*15-1)35+15^2-3*15+2}{2}=1211

Same:

For (2): y \ number \ position=\dfrac{15^2+(2*35-3)15+35^2-35+2}{2}=1211

I had done a 'programme', showing how works this topic, but I have not been able to attach the file of Excel.(Headbang)Regards.

 

[chisigma]

Hi MHB,

My nephew, age 9, was asked the following question and he hoped I could solve the problem and then explain the solution to him using only elementary math concepts. My boyfriend has solved it, but he used a formula that he recalled seeing in a textbook by G.H. Hardy, and that involved knowledge of triangular numbers.

So, I don't know how to attack this problem from a 9 year old's perspective. If you happen to know how to simplify the problem, could you please enlighten me? Many thanks in advance.

Problem:
In the sequence of fractions $\dfrac{1}{1},\,\dfrac{2}{1},\,\dfrac{1}{2},\, \dfrac{3}{1},\,\dfrac{2}{2},\,\dfrac{1}{3},\, \dfrac{4}{1},\,\dfrac{3}{2},\,\dfrac{2}{3},\, \dfrac{1}{4},\, \dfrac{5}{1},\, \dfrac{4}{2},\, \dfrac{3}{3},\, \dfrac{2}{4},\,\dfrac{1}{5}\cdots$, fractions equivalent to any given fraction occur many times. For example, fractions equivalent to $\dfrac{1}{2}$ occur for the first two times in positions 3 and 14. In what position is the fifth occurrence of a fraction equivalent to $\dfrac{3}{7}$?

Let's suppose to call $a_{n}$ the sequence at numerator and $b_{n}$ the sequence at denominator. It will be $a_{n}=3$ for $n = 4\ m + \frac {(m-1)\ (m-2)}{2}, m=1,2,...$ and $b_{n}=7$ for $k = 7\ (k+3) + \frac{(k-1)\ (k-2)}{2}, k=1,2,...$. By systematic search You find that is $a_{n}=3,\ b_{n} = 7$ for $k=3,\ m=7, n=43 $ and no other values of n. The succcesive search is dedicated to find the value of n for which is $a_{n}=6,\ b_{n}= 14$ and then the value of n for which $a_{n}=9,\ b_{n} = 21$ and so on, so that the solution of the problem is the value of n for which $a_{n}=15,\ b_{n}=35$. That is a little hard task for an old wolf like me but I'm sure that You will find the number n with no difficulties (Wink)...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Hi MHB,

My nephew, age 9, was asked the following question and he hoped I could solve the problem and then explain the solution to him using only elementary math concepts. My boyfriend has solved it, but he used a formula that he recalled seeing in a textbook by G.H. Hardy, and that involved knowledge of triangular numbers.

So, I don't know how to attack this problem from a 9 year old's perspective. If you happen to know how to simplify the problem, could you please enlighten me? Many thanks in advance.

Problem:
In the sequence of fractions $\dfrac{1}{1},\,\dfrac{2}{1},\,\dfrac{1}{2},\, \dfrac{3}{1},\,\dfrac{2}{2},\,\dfrac{1}{3},\, \dfrac{4}{1},\,\dfrac{3}{2},\,\dfrac{2}{3},\, \dfrac{1}{4},\, \dfrac{5}{1},\, \dfrac{4}{2},\, \dfrac{3}{3},\, \dfrac{2}{4},\,\dfrac{1}{5}\cdots$, fractions equivalent to any given fraction occur many times. For example, fractions equivalent to $\dfrac{1}{2}$ occur for the first two times in positions 3 and 14. In what position is the fifth occurrence of a fraction equivalent to $\dfrac{3}{7}$?

I think this is a hard problem for a nine-year-old. The first step is to see that the fifth occurrence of a fraction equivalent to $\dfrac{3}{7}$ is the fraction$\dfrac{3\times5}{7\times5} = \dfrac{15}{35}.$

The given sequence consists of blocks of terms in which the sum of the numerator and denominator is constant. The series starts with a single term for which that sum is $2$. Then there is a block of two terms where the sum is $3$, a block of three terms where the sum is $4$, and so on. Within each block, the terms are ordered by their denominator. The first term in the block has denominator $1$, then $2$, and so on.

For the fraction $\dfrac{15}{35}$, the sum of the numerator and denominator is $15+35 = 50$, so it lies in the $49$th block. It will be the $35$th term in that block, so it will be $35$ places beyond the end of the $48$th block.

At this stage, I think that you are inevitably going to get drawn into the realm of triangular numbers. Somehow or other, you have to convince yourself that the $48$th block ends at the term whose position in the sequence is $\frac12\times 48\times 49 = 1176.$ It then follows that the fraction $\dfrac{15}{35}$ occurs at place $1176 + 35 = 1211.$

The average nine-year-old would probably find that argument too difficult to follow. But I am sure that a relative of anemone would relish it. (Wink)

Edit. To convince someone inexperienced in algebra to believe the formula $\frac12n(n+1)$ for the $n$th triangular number, I would draw an $n\times(n+1)$ array like this:

 + - - - - -
 + + - - - -
 + + + - - -
 + + + + - -
 + + + + + -

You can see from this that half of the $n(n+1)$ elements are $+$s, and also that the number of $+$s is $1+2+3+\ldots + n$.

 

[anemone]

Hi mente oscura and chisigma,

Thanks for your replies and I understand what both of you are saying, but for this problem, I have to solve it using the simplest way possible since it is a problem for my nephew, not me...if it were up to me to solve it, then I would attack it as I would attack a ferocious tiger in the forest, hehehe...:)

Hi Opalg,

Thank you so so much for the easy to follow explanation! I appreciate it from the bottom of my heart. I will also tell my nephew who Opalg is and also let him know you are someone I hold in the highest esteem. (Sun)

You can be sure that I am your number one "fan" on earth! (heart)