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

(Solution: Solution 2)
m (Solution 2)
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=> <math>n = 15384</math>.
 
=> <math>n = 15384</math>.
  
=> The original number = <math>153846</math>.
+
=> 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}}

Revision as of 00:51, 19 February 2022

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 $n$.

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

Let the original number = $10n + 6$, where $n$ is a 5 digit number.

Then we have $4(10n + 6) = 600000 + n$.

=> $40n + 24 = 600000 + n$.

=> $39n = 599976$.

=> $n = 15384$.

=> The original number = $\boxed{153\,846}$.

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