Let x be an n digit number. then each digit will appear exactly (n-1) times in tens place and (n-1) in ones place. So for any digit d we have to add d*11*(n-1). Summing for all digits we have x = S(x)*11*(n-1), where S(x) is sum of digits. 132 = (1+3+2)*11*(3-1)
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OCD attack! It should be 12 + 13 + 21+ 23 + 31+ 32. In that exact sequences!
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132 is also the address on KermitStrasse where Dieter Knausshoff (1859-1932) designed the worlds first schnitzel cloning device capable of creating 453 schnitzels per day. He is best remembered for his words, “Was mache ich mit all diesen verdammten Schnitzeln?”
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Here is more of such numbers, but considering sum of k-digit numbers (k<20). There is much more of them, but they don't fit. I suspect there is infitely many of possible k (checked up k=80). Also for given k I list all of such numbers.pic.twitter.com/5KiQBJ4Ajp
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Unfortunately there will only be finitely many such numbers: numbers grow exponentially in the number of digits that they have. But summing pairs of digits only allows for quadratic growth, so after some point the pairs won’t be able to “keep up.”
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Then the question is what is the largest of such numbers?
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11 is smaller.
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11 doesn't work, you have to count both permutations. Any 2 digit number won't work for the same reason.
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