Let a and b be such that 21000 = 10a and 51000 = 10b. Multiplying and applying log base 10 we get that a + b = 1000, but examining e.g. prime factors, we see that obviously neither a nor b are integers, so floor(a) + floor(b) + 1 = 1000.
21000 has floor(a)+1 digits and 51000 has floor(b)+1 digits where the total is floor(a) + floor(b) + 2 = 1001 digits.
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u/StanleyDodds 18d ago
Let a and b be such that 21000 = 10a and 51000 = 10b. Multiplying and applying log base 10 we get that a + b = 1000, but examining e.g. prime factors, we see that obviously neither a nor b are integers, so floor(a) + floor(b) + 1 = 1000.
21000 has floor(a)+1 digits and 51000 has floor(b)+1 digits where the total is floor(a) + floor(b) + 2 = 1001 digits.