Hi, could someone please explain the solution given to question 4 in here: https://docs.google.com/viewer?a=v&q=cache:hnmAe6-2b8gJ:www.bmoc.maths.org/home/ukmog-2012-solutions.pdf+&hl=en&gl=uk&pid=bl&srcid=ADGEESgwtd7UoCBq6FfZIQndvvDDehWFbBuqZ2Jr5FTX3OQQcIWjxGNz_lkvmR-kBK_SXSBG2eusRoiWRdg8dPOFuTdOQ2VYG1q5XDO2Hj5YgqhkSe1y3404R0LazKRqjZXl9fda5MmY&sig=AHIEtbS5YjfAO6mxRl0UsvfZRtiUAsMUFw I understand up until the point where it states the 6

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Hi, could someone please explain the solution given to question 4 in here: https://docs.google.com/viewer?a=v&q=cache:hnmAe6-2b8gJ:www.bmoc.maths.org/home/ukmog-2012-solutions.pdf+&hl=en&gl=uk&pid=bl&srcid=ADGEESgwtd7UoCBq6FfZIQndvvDDehWFbBuqZ2Jr5FTX3OQQcIWjxGNz_lkvmR-kBK_SXSBG2eusRoiWRdg8dPOFuTdOQ2VYG1q5XDO2Hj5YgqhkSe1y3404R0LazKRqjZXl9fda5MmY&sig=AHIEtbS5YjfAO6mxRl0UsvfZRtiUAsMUFw I understand up until the point where it states the 6

Mathematics
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2^6 = 64 and 64-1 = 63, a multiple of nine.
So is it done by just trying out numbers to show that it has to be greater than 6?
Sortof, it's finding the least power of 2 that is 1 more than a multiple of 9. That shows that one number is at least that many powers of 2 greater than the other, so they can't possibly have the same number of digits.

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Ok, great thanks!
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