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?
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
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