Choose two distinct non-zero digits.
Evaluate the sum of the product and
the s.o.d. of the pair.
What is the probability that the result is the concatenation of the initially selected digits?
Example: if the digits were 4,7 the resulting 4*7+7+4=39, not the case you were after, i.e. 47 or 74
In the case of non-zero digits:
It turns out, any set of 2 digits such that at least one of the digits is a 9 will pass the above test (product of digits plus sum of digits equals the same original 2 digits in some order).
If there is no 9 then the two digits will fail.
17 = 9^2 - 8^2 (total ways minus failure ways to choose each digit)
So the Probability is 17/81 = 0.20987654320987653
It wasn't asked, but if you allow for digits to be zero, then a set of 2 digits will pass if either one or both is a 0 or a 9. So if you were to allow leading zeros, there are 10 ways to choose each digit but 8 ways to choose each digit for a failure.
10^2 - 8^2 = 36 so the
Probability of success would be 36/100 with non-zero digits included.
It is not intuitively obvious to me why s.o.d plus p.o.d. of a 2 digit number equals that number (or its reverse) if and only if its digits include a 9 (or a 0 if zeros are allowed)
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Posted by Larry
on 2022-06-17 08:30:12 |
Solution
