Adding in Rows
List any 3 numbers. It is always possible to find a subset of
adjacent numbers that add up to a multiple of 3. Can you explain
why and prove it?
Age
11 to 14
Challenge level
List any 3 numbers.
It is always possible to find a subset of adjacent numbers that add up to a multiple of 3 (that is either one, two or three numbers that are next to each other). For example:
5, 7, 1
5 and 7 are adjacent and 5 + 7 = 12 (a multiple of 3)
4, 4, 15
15 is a multiple of 3
5, 11, 2
5 + 11 + 2 = 18 (a multiple of 3)
Can you explain why and prove it?
What happens if you write a list of 4 numbers?
Is it always possible to find a subset of adjacent numbers that add up to a multiple of 4?
Can you explain why and prove it?
What happens if you write a long list of numbers (say n numbers)?
Is it always possible to find a subset of adjacent numbers that add up to a multiple of $n$?
Can you explain why and prove it?
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 $a$: $${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over \displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over \displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$
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 $a$: $${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over \displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over \displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$