Step 1: Define the Number n
Let \(n\) be a number that ends in \(6\), meaning we can express it as:
\(n=10m+6n\)
where \(m\) represents the number formed by the digits of \(n\) excluding the last digit \((6)\).
Step 2: Rearranging the Digits
When we move the last digit \((6)\)to the front, the new number becomes:
\(6×10^k+m\)
where \(k\) is the number of digits in \(m\).
According to the problem statement, this new number is four times the original number
\(n\) : \(6×10^k+m=4n\)
Substituting \(n=10m+6n\) into the equation:
\(6×10^k+m=4(10m+6)\)
Expanding:
\(6×10^k+m=40m+24\)
Rearrange:
\(6×10^k−24=39m\)
\(3(2×10^k-8)=3(13m)\)
Dividing by 3:
\(2×10^k−8=13m\)
\(2×10^k=13m+8\)
Step 3: Find the Smallest Natural n
To find the smallest \(n\), we need to determine the smallest \(k\) such that:
\(2×10^k−8\) is divisible by \(13\).
Checking values for \(k\):
- For k=2: \(2×10^2−8=200−8=192\)
\(192\) is not divisible by \(13\). - For k=3: \(2×10^3−8=2000−8=1992\)
\(1992÷13=153.23\), not an integer. - For k=4: \(2×10^4−8=20000−8=19992\)
\(19992÷13=1538\), which is an integer.
So, \(m=1538\) and \(n=10m+6=10×1538+6=15386\)
Step 4: Verify the Solution
Moving the last digit \(6\) to the front:
\(6×10^4+1538=60000+1538=61538\)
Checking if this equals \(4n\):
4×15386=61544
Since \(61538≠61544\), Let’s check the next value of \(k\).
- For k=5: \(2×10^5−8=200000−8=199992\)
\(199992÷13=15384\), which is an integer.
\(n=10×15384+6=153846\)
Verifying: \(6×10^5+15384=600000+15384=615384\)
\(4×153846=615384\)
Since the condition holds, the smallest number \(n\) is:
\({153846}\)
Final Answer
153846
Also Read About : Solution for the IMO 2017 Problem: Finding All Prime Triples (p,q,r) Satisfying p^q+q^p=r
Khairun
