What is the smallest number so that if you move its last digit to the beginning (for example, turning 1234 into 4123) you get a new number that is an integer multiple of the original number?

The required number has to be in the form of X=A1A2A3...An. By moving An to the front, the number becomes Y = AnA1A2A3...A(n-1). From the problem statement we get: Y = kX where k is an integer between 2 and 9. Also to eliminate trivial solutions, A1 has to be greater than 0 and An has to be greater than 1.

Therefore, X/10^n = 0.A1A2A3...An, and

Y/10^n = 0.AnA1A2A3...A(n-1)=(kX)/10^n

Also, to get a repeating decimal you have to divide by 10^n-1. Therefore, let p = X/(10^n-1)  = 0.A1A2A3...AnA1A2A3..., and let q = Y/(10^n-1) = 0.AnA1A2A3...A(n-1)AnA1A2A3... = (kX)/(10^n-1) = kp

p/10 = 0.0A1A2A3...A(n-1)AnA1A2A3..., and

An/10 = 0.An

p/10 + An/10 = 0.AnA1A2A3...A(n-1)AnA1A2A3...

Therefore, q = p/10 + An/10

10q = p + An

we also have, q = kp

Substituting gives: 10kp = p + An

10kp - p = An

p(10k - 1) = An

p = An/(10k - 1) .................................(Eq. 1)

From the problem statement, we can deduce the following:

9 <= An and An>= k

From equation 1: p is minimum when An = k

Substituting the integers 2 through 9 for An and k in equation 1 yields the following repeating decimals:

2: 2/19 = 0.10526315789...

3: 3/29 = 0.10344827586...

4: 4/39 = 0.102564... (repeating)

5: 5/49 = 0.10204081632...

6: 6/59 = 0.10169491525...

7: 7/69 = 0.10144927536...

8: 8/79 = 0.10126582278...

9: 9/89 = 0.10112359550...

From the above results 4 has the fewest repeating digits, Therefore An = k = 4 and the smallest number is: 102564

 

Candide

 

Substitue

Posted by Candide on 2004-09-28 17:04:01