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

See Also

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