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Helen's Conjecture

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Age 11 to 14 Challenge Level:

Several nice solutions were sent in- a sample of which are listed below as they differ slightly. You might like to decide which one makes most sense to you.

Many of you realised that each recurring decimal was ten times as great as the previous one but could also be written as a sum of two numbers in certain cases, such as in the following

There are two ways to look at this problem

$2.22222$.... $= 10 \times0.22222$.... but it also equals $2 + 0.222222$....
$22.22222$ ....$ = 100 \times0.2222222$ ... but it also equals $22 + 0.222222$....

For the second part the fractions group up into pairs
$2.525252525$... $= 10 \times0.252525252$ but cannot be written as a nice sum.

However, $25.252525252$.... $= 100 \times0.2525252525$..... but it also equals $25 + 0.2525252525$.

This idea was used by the following students to find the form of the fraction for 2.52525252

Submitters: Michael of Spalding Grammer, Mary of Birchwood Community High School, Andrei of School 205, Bucharest, Tan Chor or the Raffles Institution, Sana, Jenny, Chris and Rosion of Madras College, St. Andrews and Chen of The Chinese High School, Singapore .


Let $2.5252\ldots$ be $x$

$x = 2.5252\ldots$
$100x = 252.5252\ldots$
$99x = 100x - x$
$99x = 250$
So $x = 250 / 99$

Hence, sum of numerator and denominator $= 250 + 99 = 349$


The sum of the denominator and numerator is $349$.

We divided $5$ by $1$ which equals $5$.

We divided $5$ by $2$ which equals $2.5$ or $2 1/2$.

So we knew that $5$ had to be divided by a number between $1$ and $2$ to get $2.5252525\ldots$

Using trial and error we finally divided $5$ by $1.98$.

$5 / 1.98 = 2.52525252525252525252525\ldots$

Then we found the equivalent fractions

$5 / 1.98 = 500 / 198 = 250 / 99$

Then we added the denominator to the numerator:

$99 + 250 = 349$


Given a recurring decimal $0.xyxyxyxy\ldots $ where $x$ and $y$ represent digits, it can be shown that $0.xyxyxyxy \ldots = xy/99$

$100 * 0.xyxyxyxyxy \ldots = xy.xyxyxyxy \ldots$
Thus, $99 * 0.xyxyxyxy\ldots = xy.xyxyxyxy\ldots - 0.xyxyxyxy \ldots= xy$

Dividing $99$ from both sides, we see that $0.xyxyxyxy\ldots$ is equal to $xy/99$.

Having derived this formula, $2.525252... = 2 52/99 = 250/99$. Adding the denominator and numerator, the final answer is $349$.

Extending the above formula for all recurring decimals, it can again be derived that $0.a_1a_2a_3...a_x...$ where $a_1$ to $a_x$ represent digits that recur, is equal to $${a_1a_2a_3... a_x \over 10^x - 1}.$$