Happy Birthday Ramanujam

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Today is 127th birthday of our beloved mathematician Srinivasa Ramanujam Iyenger.
To celebrate his birthday I would like to pose a puzzle to all the readers and urge you  to solve it in your head, as Ramaujam did!!

In a certain street, there are more than fifty but less than five hundred houses in a row, numbered from 1, 2, 3 etc. consecutively. There is a house in the street, the sum of all the house numbers on the left side of which is equal to the sum of all house numbers on its right side. Find the number of this house.

[Hint: Ramanujam used concept of continued fractions]

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4 responses »

  1. If the equation you got is of form:
    \frac{n(n+1)}{2} = m^2
    then,
    if we get n, them m = \sqrt{\frac{n(n+1)}{2}} or if we get m, then n = \frac{\sqrt{1+8m^2} - 1}{2}

    Thus, this problem can be stated in two other ways.

    1. Find a triangular number (greater than 1275 but less than 125250) that is a perfect square.

    Its square root will give an m, and n can be found out by above equation.

    2. Find n (greater than 50 but less than 500) such that the sum of the first n natural numbers is a perfect square.

    This will give n, from which m can be found out using above equation.

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  2. In this problem the number of all houses will be 288 and hence the number of that particular house will be 204. See, I am not Ramanujan, so I had not solved this problem just in my head. Actually I tried it but just then, I used a calculator.

    Well, if you consider the general form of this problem, then the total number of houses in the street will must be a certain specific number.The first ten possible numbers for total numbers of houses are:
    1, 8, 288, 332928, 443365544448, 786292024016459316676608, 2473020588127600939387543243786675530709484249088, 24463323317211940977293404419928347279246091129334268572773850449273611455484253634058690852323328,
    2393816750889781775169375739082176313196905492596202627327114638125962681393885153176192162692298199297165620016318855149608788834131565123446110899404713627587174222798551817391907017545221275648,
    22921434547362046141128332279762608346086545581263248509428408603726883563547677026209036319928850444345567180819748621885764088358391438228874398472642124797435100191960402661073499888961891659753863641925211415654957803888907383474079644981677759147432829941948356068250928002306759056275245275522840370922866791882602874296650381112638853121485359910707340030101733739771961875110476382208.

    Also, one can find that there are infinitely many numbers for the total number of houses in the street.

    Actually, if a number ‘x’ denotes the total possible numbers of houses in the street, then the number ‘4x(x+1)’ will give another number for total numbers of houses. Starting from number ‘x=1’, we can get all possible numbers for the total number of houses in the street.

    Now, for each ‘x’, \sqrt{\frac{x \times(x+1)}{2}} will give the number of house for which, the sum of all the house numbers on the left side of will equal to the sum of all house numbers on its right side.

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    • This problem was based on “Pell’s Equation” .
      So you have discovered a way to solve such Diophantine equations. Congratulations !!

      NOTE: In Number Theory, one is allowed to use only simple calculators (not computers) to experiment with conjecture.

      Like

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