anonymous
  • anonymous
Let A be a number with 2001 digits such that A is a multiple of 10! Let B be the digit sum of A , C be the digit sum of B , and D be the digit sum of C . What is the unit’s digit of D ? any help appreciated
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Its a good question isnt it?
shubhamsrg
  • shubhamsrg
yes, it is.
KingGeorge
  • KingGeorge
Well, I don't have a rigorous way to show that this is true for all multiples of 10! with 2001 digits, but I can do it for certain examples. We can calculate 10!=3628800 By multiplying this by 10 repeatedly, we can obtain a number 36288000...0. This, when you sum the digits, gets you 9 (which is good since 9 divides 10!). But you get that as soon as you calculate C. To show this rigorously, you would need to show that as soon as you calculate D, you get 9. As a side note, it's easy to show that you eventually get 9, but it may take more than 3 digit sums

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KingGeorge
  • KingGeorge
Well, if we assume the worst case, and 999...9 (2001 times) is actually a multiple of 10!, then we get that the digit sum is 18009. So any digit sum is no more than 5 digits long. Again, at worst these are all 9's, so the digit sum is 45. This has a digit sum of 9. So after 3 digit sums, any 2001 digit number will only be a single digit. Thus, the unit's digit of D is 9.

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