Proving that 31123319 is the largest number with a self-accounting property is 10213223. What is the largest such number?

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undodaonePvopxl24 · Accepted Answer

Step 1A method of attack - illustrated for 8 digit numbers, where no particular digit occurs more than 9 times. Initially I will assume that the order of the pairs is irrelevant.Let the number be aAbBcCdD. Then a+b+c+d=8 with each of a,b,c,d being non-zero.For the 8 digit number to have initial digit at least 3, the possibilities for {a,b,c,d} are {5,1,1,1}, {4,2,1,1}, {3,3,1,1} or {3,2,2,1}Now apply the self-accounting property to each possibility:{5,1,1,1}={A,A,A,A} Impossible{4,2,1,1}={A,A,A,B} Impossible{3,3,1,1}={A,A,B,B} Solved by A=3, B=1. Eg. 31331819{3,2,2,1}={A,A,B,C} Solved byA=2, B=3, C=1 Eg. 21322319 when the initial digit is not 3 if the counted digits are put into ascending order.

Proving that 31123319 is the largest number with

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