For any positive integer N, consider the digits which occur either inNor in 7*N. Letmbe the smallest digit among those digits. What is the largest possible value ofm?(Doh)

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- Jul 30th 2008, 02:30 AMclarebear14Largest possible value?
For any positive integer N, consider the digits which occur either in

*N*or in 7**N*. Let*m*be the smallest digit among those digits. What is the largest possible value of*m*?(Doh) - Jul 30th 2008, 06:26 AMQuick
You by far have the oddest problems presented, and I need practice because my hw is going to be basically two of these a night (Nerd)

The answer is 6.

here's the proof:

2*7 = 14 so any number that starts with 2, will have a 1 carried over to the next place. Example: 28394*7 = 198758 <-- 7*N will always start with 1 if N starts with 2

3*7 = 21 so if N starts with 3 then 7*N starts with 2

here's the whole list:

2 --> 1

3 --> 2

4 --> 2

5 --> 3

6 --> 4

7 --> 4

8 --> 5

9 --> 6

Now it's important to note here that there are two exceptions to this rule, when N starts with 1 then it can be anything, because numbers can carry over, but if N starts with 1 then m would be 1 so no possible value of N starts with 1.

The second is that there can be a number which carries over. However, because that number would be the first digit of one one digit number times 7 and the second digit of another one digit number times 7 with possibly a one carried on top of it, the highest the second digit of 7*N could be if it sent a carry-over to the first number and the first digit is at least 7 would be: 0

That sounded really confusing I know.

Anyway, here's the deal. The highest starting number in 7*N is 7, but that would mean that there is a 0 in 7*n so m would equal 0, that means that the starting number is less than 7 so m is less than 7. It so happens that m can equal 6 and here's an example:

N = 97

7*N = 679

m = 6