Jim calculated the product of the non zero digits of each integer from 1 to 10²⁰²⁰ and then he summed these 10²⁰²⁰ products. Which number did he obtain?

Larry’s work directly leads to the explanation: 

Call Jim's product and sum taken from 1 to 10^i "S(i)". Eg, S(1)= 1+2+…..+8+9+1= 46

Consider the following method of making a list of numbers that run from 1 to 10^(i+1) from lists of numbers that run from 1 to 10^(i) - 1:  

1) Write the 10^(i)-1 list such that all numbers are strictly i digits long. E.g., for i=2, 01,02,...,99

2) For ten copies of this list, place 0,1,2,...,9 in front of the digits in each list respectively. E.g, 

001,002,…,099,  101,102,…199,  201,202,…,299, …, …, 901,902,…,999

What is Jim's count for this set of lists? They sum to 

1 (S(i)-1) + 1 (S(i)-1) + 2 (S(i)-1) + ... 9 (S(i)-1) = 46 (S(i)-1) 

But, we left out the 10 intermediate numbers that are all zero except for the leading digit. E.g, 100, 200, ..., 900, 1000, whose Jim-type sum is also 46. So, for the full list, S(i+1) = 46 (S(i) - 1) + 46 = 46 S(i).

Starting recursively from S(1)=46, we have S(i)=46^i

QED

Edited on December 27, 2020, 6:35 pm

Posted by Steven Lord on 2020-12-27 09:33:39