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36 At Right Angles | Vol. 3, No. 1, March 2014

A well-known puzzle that can be posed to even young children is as

follows:

There is a row of houses numbered sequentially from 1. A

resident of one of the houses notices one day that the sum

of the door numbers to his left is the same as the sum of

the door numbers to his right. How many houses are

there in the row and in which of these does the speaker

live?

Matters can be made simpler by stating that the number of houses

is less than 10. The solution is then not dif�icult to get by trial and

error� see if you can �ind it.

The next level of challenge is to ask for another solution to the

problem. (The numbers involved are still less than 50.) Now trial

and error may not be a good option. One could use the well-known

formula for the sum of the �irst nnnn natural numbers

SSSS� = nnnnnnnnnn nn 1nn

2

to proceed. Let nnnn be the number of houses in the row and mmmm the

door number of the house occupied by the speaker. Then the sum

of the numbers from 1 to mmmm mm 1 is equal to the sum of the numbers

from mmmm nn n to nnnn. That is,

mmmmnnmmmm mm 1nn

2 = nnnnnnnnnn nn 1nn

2 mm mmmmnnmmmm nn nnn

2 , (1)

which leads to

mmmm� = nnnnnnnnnn nn 1nn

2 . (2)

in the classroom

A Puzzle Leading

to Some Interesting

Mathematics

A. Ramachandran

A well-known puzzle that can be posed to even young children is

as follows:

There is a row of houses numbered sequentially from 1. A

resident of one of the houses notices one day that the sum of

the door numbers to his left is the same as the sum of the door

numbers to his right. How many houses are there in the row

and in which of these does the speaker live?

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At Right Angles | Vol. 3, No. 1, March 2014 37

A RAMACHANDRAN has had a long standing interest in the teaching of mathematics and science. He studied

physical science and mathematics at the undergraduate level, and shifted to life science at the postgraduate level.

He taught science, mathematics and geography to middle school students at Rishi Valley School for over two

decades, and now stays in Chennai. His other interests include the English language and Indian music. He may be

contacted at archandran.53@gmail.com.

The right hand side of the above equation

represents a triangular number, while the left

hand side represents a square number. So we look

for numbers that are both square and triangular.

Lists of such numbers are available. The lowest

such (apart from 1 itself ) is 36: it is the 6 square

number and the 8 triangular number. This

corresponds to and , and the door

number sums involved are 1 + 2 + 3 + 4 + 5 = 15

and 7 + 8 = 15. The next larger values of and

are, respectively, 35 and 49:

1 + 2 + 3 + ⋯ + 33 + 34 = 595,

36 + 37 + 38 + ⋯ + 48 + 49 = 595.

pairs of whole numbers which satisfy (2).

The table below displays some of these pairs:

1 6 35 204 1189 6930 ...

1 8 49 288 1681 9800 ... (3)

As and get larger, the ratio gets

gradually closer to √2. For example:

49

35 = 1.4,

288

204 ≈ 1.41,

1681

1189 ≈ 1.414, ... .

We see a slow approach to √2

see why this must be so: we rewrite the equation

as , from which we

get:

= 1 − ,

∴ = 1 − = 1 − 1 .

As gets large, gets close to 0, which

means that gets close to 2.

In (3) we see many striking patterns in both the

-row and the -row. Example: Write the

numbers in the -row as follows:

1 , 8 = 3 − 1, 7 ,

288 = 17 − 1, 1681 = 41 , ... . (4)

We see that there is an underlying sequence of

numbers:

1, 3, 7, 17, 41, ... , (5)

and we see that this sequence itsel� has an

underlying pattern! Each successive number can

be formed using the two previous numbers (much

like the rule for the Fibonacci numbers): if

are three successive numbers in (5), then

(6)

Example: 41 = (2 × 17) + 7. If this pattern is

valid, then we ought to be able to guess more

numbers in the sequence! After 41 we should

have (2 × 41) + 17 = 99, so the next -value after

1681 should be 99 − 1 = 98 × 100 = 9800, and

it really is so.

Let's continue the pattern. In (5), the number

following 99 should be (2 × 99) + 70 = 239, so in

(3) the number following 1681 should be

239 = 57121.

Can we now anticipate which number will come

after 6930 in the -row (the top row) of (3)?

These numbers too have a simple pattern: if

are any three consecutive numbers in the row,

then

(7)

Example: 35 = (6 × 6) − 1. Therefore after 6930

we should get:

(6 × 6930) − 1189 = 40391.

So we expect the following equality to be true:

1+2+3+⋯+40390 = 40392+40393+⋯+57121.

Indeed, it is true: both sides are equal to

815696245. (Please check! Use the fact that the

positive integers is .)

and maybe supply some proofs.

(Yes, they are needed ....)