Sure, let's solve this step-by-step.
We need to find a six-digit positive integer [tex]\( n \)[/tex] that satisfies the following conditions:
1. The last digit of [tex]\( n \)[/tex] is 6.
2. If the last digit (6) is moved to the front of [tex]\( n \)[/tex], the resulting number is 4 times [tex]\( n \)[/tex].
Let's denote [tex]\( n \)[/tex] as [tex]\( ABCDE6 \)[/tex], where [tex]\( A, B, C, D, E \)[/tex] are the other digits of the number.
When we move 6 to the front, the resulting number will be [tex]\( 6ABCDE \)[/tex].
According to the problem, [tex]\( 6ABCDE \)[/tex] must be 4 times [tex]\( ABCDE6 \)[/tex]. So, we can write the equation:
[tex]\[ 6ABCDE = 4 \times ABCDE6 \][/tex]
Let's consider the structure of these numbers:
1. If [tex]\( n = ABCDE6 \)[/tex], then it can be written as [tex]\( n = 100000A + 10000B + 1000C + 100D + 10E + 6 \)[/tex].
2. When 6 is moved to the front, the new number [tex]\( 6ABCDE \)[/tex] can be written as [tex]\( 6 \times 10^5 + 10000A + 1000B + 100C + 10D + E \)[/tex].
Now, this gives us the equation:
[tex]\[ 600000 + 10000A + 1000B + 100C + 10D + E = 4 \times (100000A + 10000B + 1000C + 100D + 10E + 6) \][/tex]
We need to use algebra to simplify and solve this equation. However, after careful consideration and solving, we find the lowest value of the original positive integer:
[tex]\[ n = 153846 \][/tex]
This is the smallest six-digit positive integer that, when its last digit (6) is moved to the front, results in a number that is 4 times the original number.
