Let $n$ be a positive integer,$s(n)$ be the sum of digits of $n$ and $t(n)$ be the sum of the stumps of $n$.
A $stump$ is a number which is obtained by removing several(at least one) digits from the right hand end of the decimal representation of $n$.
Show that $n=s(n)+9t(n)$
Stumps of n,APMO 2001-1
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Re: Stumps of n,APMO 2001-1
Let $n=\left ( n_{m}n_{m-1}......n_{0} \right )_{10}$
So\[n=\sum_{i=0}^{m}10^{i}n_{i}\]
Notice that if we vanish last $i$ digit then we will get a number $t_{i}$ in a form of\[t_{i}= \sum_{j=i}^{m}10^{j-i}n_{j}\]But, \[t(n)=\sum_{i=1}^{m}t_i\]\[\Rightarrow t(n)=\sum_{i=1}^{m}\left ( 10^{i-1}+10^{i-2}+...+1 \right )n_{i}\]\[\Rightarrow t(n)=\sum_{i=1}^{m}\left ( \frac{10^{i}-1}{9} \right )n_{i}\]Note that,\[\Rightarrow s(n)= \sum_{i=0}^{m}n_i\]So,\[\Rightarrow s(n)+9t(n)=\sum_{i=0}^{m}\left ( 10^{i}-1+1 \right )n_{i}=n\]
And we're done
So\[n=\sum_{i=0}^{m}10^{i}n_{i}\]
Notice that if we vanish last $i$ digit then we will get a number $t_{i}$ in a form of\[t_{i}= \sum_{j=i}^{m}10^{j-i}n_{j}\]But, \[t(n)=\sum_{i=1}^{m}t_i\]\[\Rightarrow t(n)=\sum_{i=1}^{m}\left ( 10^{i-1}+10^{i-2}+...+1 \right )n_{i}\]\[\Rightarrow t(n)=\sum_{i=1}^{m}\left ( \frac{10^{i}-1}{9} \right )n_{i}\]Note that,\[\Rightarrow s(n)= \sum_{i=0}^{m}n_i\]So,\[\Rightarrow s(n)+9t(n)=\sum_{i=0}^{m}\left ( 10^{i}-1+1 \right )n_{i}=n\]
And we're done
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Stumps of n,APMO 2001-1
That is correct as far I see. My solution was the same way.
One one thing is neutral in the universe, that is $0$.