1962 IMO Problems/Problem 1

Problem

Find the smallest natural number $n$ which has the following properties:

(a) Its decimal representation has 6 as the last digit.

(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number $n$ .

Video Solution

https://youtu.be/9y5UUNIhUfU?si=PzXbNokxOXCRxYBh [Video Solution by little-fermat]

Solution 1

As the new number starts with a $6$ and the old number is $1/4$ of the new number, the old number must start with a $1$ .

As the new number now starts with $61$ , the old number must start with $\lfloor 61/4\rfloor = 15$ .

We continue in this way until the process terminates with the new number $615\,384$ and the old number $n=\boxed{153\,846}$ .

Solution 2

We know from the two properties that for some string $x$ , $n=10x+6$ . Let the number of digits in $x$ be $a$ ; then moving the $6$ to the front would give it place value $10^a$ ; as a result, $4n=6\cdot10^a+x$ . Multiplying this by $10$ gives $40n=6\cdot10^{a+1}+10x$ , and subtracting the former yields $39n=6(10^{a+1}-1)$ , or $13n=2(10^{a+1}-1)$ . As a result, $13|(10^{a+1}-1)$ . By Fermat's Little Theorem, we know that $10^{12}-1$ divides $13$ , so it isn't difficult to try values of $a+1$ less than $13$ to find the smallest such $a$ .

Eventually, we notice that $10^6-1$ divides $13$ , so $a=5$ . Then $\boxed{n=153846}$ , and since the number ends in $6$ , we know that $x$ is also an integer, so this is the solution.

~eevee9406

1962 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1962 IMO Problems/Problem 1

Contents

Problem

Video Solution

Solution 1

Solution 2

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