Find the largest 7-digit number which, when divided by 5, 7, 11, and 13, respectively, leaves the same and the largest possible remainder for all four divisors.

To leave the same largest possible remainder, the number must leave a remainder of 4 when divided by 5, 6 when divided by 7, 10 when divided by 11, and 12 when divided by 13, as 4 is the largest common remainder less than all four divisors. This means the number plus 1 is divisible by 5, 7, 11, and 13, whose least common multiple is 5005. We find the largest 7-digit multiple of 5005, which is 9999945, and subtracting 1 gives 9999944, which contradicts the intended logic of the largest remainder being 6, as the number 9999994 divided by 5 leaves a remainder of 4, by 7 leaves 6, by 11 leaves 6, and by 13 leaves 6; the largest common remainder is therefore 4, making the number 9999994.

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