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

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==Solution 2==
 
==Solution 2==
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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>.
  
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Let the original number = <math>10n + 6</math>, where <math>n</math> is a 5 digit number.
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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.
  
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Then we have <math>4(10n + 6) = 600000 + n</math>.
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~eevee9406
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=> <math>40n + 24 = 600000 + n</math>.
 
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=> <math>39n = 599976</math>.
 
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=> <math>n = 15384</math>.
 
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=> The original number = <math>\boxed{153\,846}</math>.
 
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1962|before=First Question|num-a=2}}
 
{{IMO box|year=1962|before=First Question|num-a=2}}

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

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
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