Fundamental Theorem of Arithmetic Questions in English

(B) To find the prime factors of $3825$,we perform successive division by prime numbers:
$1$. Since the last digit is $5$,it is divisible by $5$: $3825 \div 5 = 765$.
$2$. $765$ is also divisible by $5$: $765 \div 5 = 153$.
$3$. Now,check for $3$: The sum of digits of $153$ is $1+5+3 = 9$,which is divisible by $3$. So,$153 \div 3 = 51$.
$4$. $51$ is divisible by $3$: $51 \div 3 = 17$.
$5$. $17$ is a prime number,so $17 \div 17 = 1$.
Thus,the prime factorization is $3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$.

(A) To express $156$ as a product of its prime factors,we perform prime factorization:
$156 = 2 \times 78$
$78 = 2 \times 39$
$39 = 3 \times 13$
Combining these,we get $156 = 2 \times 2 \times 3 \times 13$.
In exponential form,this is $2^2 \times 3 \times 13$.

(A) To determine if the decimal expansion of a rational number $\frac{p}{q}$ is terminating,we examine the prime factorization of the denominator $q$.
If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Given the rational number $\frac{13}{3125}$,the denominator is $q = 3125$.
Prime factorization of $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
This can be written as $2^0 \times 5^5$.
Since the denominator is in the form $2^n \times 5^m$ (where $n=0$ and $m=5$),the decimal expansion of $\frac{13}{3125}$ is terminating.

(B) For a number to end with the digit $0$,it must be divisible by $10$.
This means its prime factorization must contain both $2$ and $5$ as factors.
The prime factorization of $4^n$ is $(2^2)^n = 2^{2n}$.
Since the only prime factor of $4^n$ is $2$,it does not contain $5$ as a factor.
Therefore,$4^n$ can never end with the digit $0$ for any positive integer $n$.