Answer :
To determine whether [tex]\( 15^n \)[/tex] can end with the digit zero for any natural number [tex]\( n \)[/tex], we need to analyze the properties of numbers that end in zero. A number ends in zero if it is divisible by 10.
Mathematically, we can state this as:
[tex]\[ 15^n \equiv 0 \pmod{10} \][/tex]
For [tex]\( 15^n \)[/tex] to be congruent to 0 modulo 10, [tex]\( 15^n \)[/tex] needs to be divisible by 10. We know that a number is divisible by 10 if it has both 2 and 5 as factors (since [tex]\( 10 = 2 \times 5 \)[/tex]).
Let's consider [tex]\( 15^n \)[/tex]:
[tex]\[ 15 = 3 \times 5 \][/tex]
So,
[tex]\[ 15^n = (3 \times 5)^n = 3^n \times 5^n \][/tex]
We now need [tex]\( 3^n \times 5^n \)[/tex] to be divisible by 10. For this product to include both a factor of 2 and a factor of 5, it is required that both [tex]\( 3^n \)[/tex] and [tex]\( 5^n \)[/tex] contribute to making the product divisible by 10.
Examining the factors:
- [tex]\( 5^n \)[/tex] certainly has [tex]\( 5 \)[/tex] as one of its factors.
- However, [tex]\( 3^n \)[/tex] does not contribute a factor of 2 because [tex]\( 3 \)[/tex] and any power of [tex]\( 3 \)[/tex] lacks the prime factor [tex]\( 2 \)[/tex].
Since [tex]\( 3^n \)[/tex] is never divisible by 2, the product [tex]\( 3^n \times 5^n \)[/tex] will also never be divisible by 2. As a result, without a factor of 2, [tex]\( 15^n \)[/tex] cannot be divisible by 10.
Therefore, [tex]\( 15^n \)[/tex] can never end with the digit zero, regardless of the value of [tex]\( n \)[/tex]. Thus, [tex]\( 15^n \)[/tex] does not end with the digit zero for any natural number [tex]\( n \)[/tex].
Mathematically, we can state this as:
[tex]\[ 15^n \equiv 0 \pmod{10} \][/tex]
For [tex]\( 15^n \)[/tex] to be congruent to 0 modulo 10, [tex]\( 15^n \)[/tex] needs to be divisible by 10. We know that a number is divisible by 10 if it has both 2 and 5 as factors (since [tex]\( 10 = 2 \times 5 \)[/tex]).
Let's consider [tex]\( 15^n \)[/tex]:
[tex]\[ 15 = 3 \times 5 \][/tex]
So,
[tex]\[ 15^n = (3 \times 5)^n = 3^n \times 5^n \][/tex]
We now need [tex]\( 3^n \times 5^n \)[/tex] to be divisible by 10. For this product to include both a factor of 2 and a factor of 5, it is required that both [tex]\( 3^n \)[/tex] and [tex]\( 5^n \)[/tex] contribute to making the product divisible by 10.
Examining the factors:
- [tex]\( 5^n \)[/tex] certainly has [tex]\( 5 \)[/tex] as one of its factors.
- However, [tex]\( 3^n \)[/tex] does not contribute a factor of 2 because [tex]\( 3 \)[/tex] and any power of [tex]\( 3 \)[/tex] lacks the prime factor [tex]\( 2 \)[/tex].
Since [tex]\( 3^n \)[/tex] is never divisible by 2, the product [tex]\( 3^n \times 5^n \)[/tex] will also never be divisible by 2. As a result, without a factor of 2, [tex]\( 15^n \)[/tex] cannot be divisible by 10.
Therefore, [tex]\( 15^n \)[/tex] can never end with the digit zero, regardless of the value of [tex]\( n \)[/tex]. Thus, [tex]\( 15^n \)[/tex] does not end with the digit zero for any natural number [tex]\( n \)[/tex].
