What is the largest multiple of 55 using the digits 0, 1, 2, 3, 4, 5, 6 without repetition?
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Call this number N . We need N to divide 5 , therefore the last digit must equal 0 or 5 . Furthermore, we need N to divide 1 1 , which means that the sum of the odd digits and the sum of the even digits must be equal mod 1 1 .
The total sum of the digits is of course 0 + 1 + 2 + 3 + 4 + 5 + 6 = 2 1 : an odd number, therefore the 'odd sum' (call it O ) and 'even sum' (call it E ) cannot be precisely equal. The only remaining possibility is that they differ by 1 1 , meaning one is equal to 5 and the other to 1 6 . O is the sum of 4 different digits, meaning O ≥ 6 . Therefore O = 1 6 and E = 5 .
To make N as large as possible, we can take its first digit to be 6 . The second digit cannot be equal to 5 , seeing as it is one of three odd digits that sum to 5 . We can take it to be 4 , forcing the other odd digits to be 1 and 0 . Therefore the last digit must be equal to 5 . Arranging the other digits from high to low (to make N as large as possible), we obtain N = 6 4 3 1 2 0 5 .