2017 UNCO Math Contest II Problems/Problem 4

Problem

Monkey Business

Harold writes an integer; its right-most digit is 4. When Curious George moves that digit to the far left, the new number is four times the integer that Harold wrote. What is the smallest possible positive integer that Harold could have written?

Solution

Solution 1

Let $x=0.\overline{a_1a_2\ldots a_n4}$ . When we multiply this number by $4$ , we have $4x=0.\overline{4a_1a_2\ldots a_n}$ . Multiplying by $10$ , $40x=4.\overline{a_1a_2\ldots a_n4}$ , which is precisely $4+x$ . Thus we have $x=\frac{4}{39}=0.\overline{102564}$ , so the answer is \boxed{102564}

Solution 2

Let $n$ denote the number of digits in Harold's integer and x be the remaining part. Then we have the equation $4\cdot 10^n+x=10x+4$ Which simplifies to $4\cdot (10^n-1)=39x$ This means that $10^n-1$ has to be divisible by $39$ (since $4$ is relatively prime to $39$ ). But this just tells us that $n$ is the length of the period of $1/39$ and $x$ is just $4$ times one period of $1/39$ , or one period of $4/39$ . But since $4/39$ is $0.\overline{102564}$ , so our answer is $\boxed{102564}$

Solution 3

We proceed with basic multiplication

_ _ _ _ _ 4
×         4
———————————
4 _ _ _ _ _

4 × 4 = 16
        1
_ _ _ _ 6 4
×         4
———————————
4 _ _ _ _ 6

4 × 6 = 24
      2 1
_ _ _ 5 6 4
×         4
———————————
4 _ _ _ 5 6

4 × 5 = 20
    2 2 1
_ _ 2 5 6 4
×         4
———————————
4 _ _ 2 5 6

4 × 2 = 8
  1 2 2 1
_ 0 2 5 6 4
×         4
———————————
4 _ 0 2 5 6

4 × 0 = 0
  1 2 2 1
1 0 2 5 6 4
×         4
———————————
4 1 0 2 5 6

4 × 1 = 4

So the minimum value is $\boxed{102564}$

2017 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions

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