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?
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Originally posted by luskinyes, a good math magic trick
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 ThomasterLet SOD be the function that, with any number as input, gives the Sum Of Digits of that number.
Do you have a proof for it?
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.
Originally posted by TheMaster37nice. 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.)
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.