SOLUTION: There is a number that is called N. When N has a seven in front of it, the number formed is five times as large as when there is a seven after it. What is the smallest number of di
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Question 1131572: There is a number that is called N. When N has a seven in front of it, the number formed is five times as large as when there is a seven after it. What is the smallest number of digits in N? Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52810) (Show Source):
From the condition, N is the minimal whole number satisfying the equation
= (1)
for some integer positive "n". It is equivalent to
= 49N.
So, we need to find "n" in a way is a multiple of 49.
The "trying and error" method, facilitated with Excel, quickly gives such a minimal N = 14285, n = 5.
Check. Left side of eq(1) is = 714285.
Right side of eq(1) is 5*(10*14285+7) = 714285. ! Correct !
ANSWER. The number is 14285.
The solution by tutor @ikleyn is, as usual, excellent; and finding the solution using an excel spreadsheet is a way to find the answer quickly and easily.
I will show a couple of ways to get the answer without a spreadsheet.
The beginning of my path to the solution is the same as hers, but it quickly diverges....
Here the task is to find the smallest number of the form (10^n)-5 that is a multiple of 7. We can do that with modular arithmetic.
Numbers of the form 10^n-5 are of the form 99...995. We want to find the smallest number of that form that is divisible by 7. Of course we could just try 995/7, 9995/7, ... to find the answer; but let's see how we can find the answer using modular arithmetic.
I will use "==" to indicate congruence mod 7....
5 == 5 mod 7
90 == -1 mod 7 (so 95 = 90+5 == 4 mod 7)
900 == -3 mod 7 (so 995 = 900+90+5 == 1 mod 7)
9000 == -2 mod 7 (so 9995 = 9000+900+90+5 == -1 mod 7)
90000 == 1 mod 7 (so 99995 = 90000+9000+900+90+5 == 0 mod 7)
99995 is the smallest number of the required form that is divisible by 7, so
Yes; that is a solution method that requires a lot of time. But it is a good demonstration of the power of modular arithmetic as a mathematical tool.
And now for what is by far the easiest way to solve this problem. (But only a real math nerd like me will recognize it.)
See that the repeating decimal for 5/7, .714285, is 5 times the repeating decimal for 1/7, .142857; and those two repeating decimals are ".14285" with 7 at either end....
