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Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

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Adding in Rows

Age 11 to 14
Challenge Level
Congratulations to Soh Yong Sheng from Raffles Institution, Singapore for this excellent solution.

We have $0 < a < b$ which means $1/a > 1/b$ and so $3 + 1/a > 3 + 1/b$. Flipping over the fraction, we will get 1/(3 + 1/a) < 1/(3 + 1/b) and the inequality remains the same way round when 2 is added. Flipping over again for the last time we get \[ \frac{1}{2+\frac{1}{3+\frac{1}{a}}} \] is greater than \[ \frac{1}{2+\frac{1}{3+\frac{1}{b}}} \]

The second part is a further expansion of the first, and in the process of repeating the above we know that it involves just one more flipping over of the fraction, thus \[ \frac{1}{2+\frac{1}{3+\frac{1}{4 + \frac{1}{a}}}} \] is less than the same thing with $b$ in place of $a$ as the inequality would be reversed again.

Lastly the continued fractions are expanded all the way down to $100 + 1/a$ and $100 + 1/b$. Observe the above process, we can tell that if the last or biggest number is odd then the continued fraction with a in it is bigger. If the last or biggest number is even then the continued fraction with $b$ in it is bigger. Each successive continued fraction involves one more 'flipping over' and reverses the inequality one more time. The following continued fraction is smaller than the same thing with $b$ in place of $a$: $${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over \displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over \displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$