We just need to show that for each of the digits in n, the sum S(n)=9T(n) contributes the value of this digit.
Consider the ith decimal digit from the left ai. The value it contribute to n is ai*10^(k-i)
It contributes ai to the sum of S(n)
The ai contributes only to the trunks of i digits or more (i≤t≤k-1):
Specifically
for t=i it contributes ai
for t=i+1 it contributes ai*10
and so on up to the largest trunk where
for t=k-1 it contributes ai*10^(k-1-i)
The total contribution to T(n) is then
ai*(1+10+100+...+10^(k-1-i))
And finally the total contribution to S(n)+9T(n) is
ai+9ai(1+10+100+...+10^(k-1-i)) = ai*10^(k-i)
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Posted by Jer
on 2021-01-04 14:10:22 |
Solution
