I will pick a number, an integer of say, 6 or more digits. Now I'll make another number by re-arranging those same digits to a different permutation of the original. Then subtract the smaller from the larger number. and from the difference, I'll select one non-zero digit to keep secret. If I reveal the other digits of the difference, in no particular order, how would you determine the secret digit?
Originally posted by doodinthemood I'm thinking you add up all the digits you tell me, then cast out nines and give the value that makes it add up to 9.
IE. You say "4,5,2,6,4" and I think "4+5 = 9, + 2 = 2, + 6 = 8, + 4 = 9 + 3 = 3. So the missing digit is 6"
Originally posted by luskin I will pick a number, an integer of say, 6 or more digits. Now I'll make another number by re-arranging those same digits to a different permutation of the original. Then subtract the smaller from the larger number. and from the difference, I'll select one non-zero digit to keep secret. If I reveal the other digits of the difference, in no particular order, how would you determine the secret digit?
Originally posted by doodinthemood I'm thinking you add up all the digits you tell me, then cast out nines and give the value that makes it add up to 9.
IE. You say "4,5,2,6,4" and I think "4+5 = 9, + 2 = 2, + 6 = 8, + 4 = 9 + 3 = 3. So the missing digit is 6"
Originally posted by TheMaster37 Let SOD be the function that, with any number as input, gives the Sum Of Digits of that number.
To prove: SOD(A) is dividable by 9 i.a.o.i. A is dividable by 9
Proof:
Let A = a*1 + b*10 + c*100 + d*1000 + ....
A mod 9 = a*1 + b*1 + c*1 + d*1 +... mod 9 = SOD(A) mod 9
Thus, if 9 divides A then 9 also divides SOD(A) and vise versa.
nice. this same modulo arithmetic argument works for proving most of the classic divisibility rules we all know and love. (taking the digit sum and checking for divisibility by 3 or 9... etc.)
24 Oct '08 05:40
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