SMO Junior Round 1 2014 #1

The digits add up to $21$ and the sums of the digits in the odd and even places have to differ by a multiple of $11$ , so they are $5$ and $16$ (can't be $16$ and $5$ because the three odd digits can add up to at most $6+5+4=15$ ). The odd digits have to therefore be $0,1,4$ or $0,2,3$ (these are the only combinations of three digits that sum to $5$ ). Since the number ends in $0$ or $5$ , it has to end in $5$ . Any number satisfying these conditions (digits are $0,1,2,3,4,5,6$ , digits in odd places are $0,1,4$ or $0,2,3$ , ends in $5$ ) is divisible by $55$ .

To maximize the number, the odd and even digits should be in descending order given the conditions. So either $6431205$ or $6342105$ , depending on your choice of the odd digits. The first one, $\fbox{6431205}$ , is larger.

A number that is divisible by 55 has to be divisible by 5 and 11. Hence, the number has to end with a 0 or 5 and have a difference of 0 or 11 between the odd digits and even digits of the number.

From here, I guess you can manually solve it.

However, there is another piece of important information that would allow you to solve it faster.

You notice that the number comprises of 7 digits and that the digits are in consecutive order. This would mean that there would be an extra one on either side of the sum of the even digits or the sum of the odd digits if you try to subtract. (Not sure if I am using the appropriate phrasing to say what I want to say)

From here, you can deduce that the number cannot end with a 5 and has to end with a 0 and that the difference between the odd and even digits has to be 11.

From here, I cannot seem to find any more clues to help solve this question so I guess you need to solve manually from here.

Therefore, you will get 6431205 as the answer.

Anyone has a better solution?