Difference between revisions of "1962 IMO Problems/Problem 1"

Latest revision as of 10:09, 15 March 2025

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

@@ Line 6: / Line 6: @@
 (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 <math>n</math>.
-==Solution==
+==Video Solution==
-{{solution}}
+https://youtu.be/9y5UUNIhUfU?si=PzXbNokxOXCRxYBh
+[Video Solution by little-fermat]
+==Solution 1==
+As the new number starts with a <math>6</math> and the old number is <math>1/4</math> of the new number, the old number must start with a <math>1</math>.
+As the new number now starts with <math>61</math>, the old number must start with <math>\lfloor 61/4\rfloor = 15</math>.
+We continue in this way until the process terminates with the new number <math>615\,384</math> and the old number <math>n=\boxed{153\,846}</math>.
+==Solution 2==
+We know from the two properties that for some string <math>x</math>, <math>n=10x+6</math>. Let the number of digits in <math>x</math> be <math>a</math>; then moving the <math>6</math> to the front would give it place value <math>10^a</math>; as a result, <math>4n=6\cdot10^a+x</math>. Multiplying this by <math>10</math> gives <math>40n=6\cdot10^{a+1}+10x</math>, and subtracting the former yields <math>39n=6(10^{a+1}-1)</math>, or <math>13n=2(10^{a+1}-1)</math>. As a result, <math>13|(10^{a+1}-1)</math>. By [[Fermat's Little Theorem]], we know that <math>10^{12}-1</math> divides <math>13</math>, so it isn't difficult to try values of <math>a+1</math> less than <math>13</math> to find the smallest such <math>a</math>.
+Eventually, we notice that <math>10^6-1</math> divides <math>13</math>, so <math>a=5</math>. Then <math>\boxed{n=153846}</math>, and since the number ends in <math>6</math>, we know that <math>x</math> is also an integer, so this is the solution.
+~eevee9406
 ==See Also==
 {{IMO box|year=1962|before=First Question|num-a=2}}

1962 IMO (Problems) • Resources
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