Difference between revisions of "1962 IMO Problems/Problem 1"
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(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>. | (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== |
| + | |||
| + | 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 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>. | ||
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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>. | 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== | ||
| + | |||
| + | Let the original number = <math>10n + 6</math>, where <math>n</math> is a 5 digit number. | ||
| + | |||
| + | Then we have <math>4(10n + 6) = 600000 + n</math>. | ||
| + | |||
| + | => <math>40n + 24 = 600000 + n</math>. | ||
| + | |||
| + | => <math>39n = 599976</math>. | ||
| + | |||
| + | => <math>n = 15384</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}} | ||
Latest revision as of 23:36, 3 September 2023
Problem
Find the smallest natural number 
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 .
Video Solution
https://youtu.be/9y5UUNIhUfU?si=PzXbNokxOXCRxYBh [Video Solution by little-fermat]
Solution 1
As the new number starts with a 
and the old number is 
of the new number, the old number must start with a 
.
As the new number now starts with 
, the old number must start with 
.
We continue in this way until the process terminates with the new number 
and the old number 
.
Solution 2
Let the original number = 
, where is a 5 digit number.
Then we have 
.
=> 
.
=> 
.
=> 
.
=> The original number = 
.
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 | ||
