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Maybe you'd lop off the last few digits to make it easier: Let A be the highest digit, B be the lowest, and X be the ones in the middle. Find X, A, and B, where
by byrneseyeview 18y ago
Maybe you'd lop off the last few digits to make it easier: Let A be the highest digit, B be the lowest, and X be the ones in the middle. Find X, A, and B, where 2(X-A*E16+A+BE16-B)=X
That way, you can at least get each digit on one side, which might make it a more tractable problem. This is very frustrating! It's easy to narrow it down a whole lot, but it's still going to be intractable.
- shrughes 18y agoSpoiler Alert! The number in general (I said, Spoiler Alert!) can be written as _y_ _x_ where 0 <= x <= 9 and the value of the number is 10y + x. Then it must satisfy for some k 2(10y+x) = (10^k)x + y Simplify that and you can show that 10^k - 2 must be divisible by 19, which means k=17 (or 35, or 53, ...). Once you have that, pick x = 2, 3, or 4 and solve for y, and you have your 18-digit number. (Letting x = 1 gives you an 18-digit numbers whose first digit is 0, which doesn't count.)