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(52812)   (Show Source): You can put this solution on YOUR website!
.
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.

Solved.



Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


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.)

1/7 = .142857 repeating
2/7 = .285714 repeating
3/7 = .428571 repeating
4/7 = .571428 repeating
5/7 = .714285 repeating
6/7 = .857142 repeating

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....
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