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Find the smallest number by which 12748 be multiplied so that the product is a perfect square.
(a) 3187
(b) 7814
(c) 4577
(d) 1842

Last updated date: 18th Jun 2024
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Hint: We arrange the numbers from given in the option in ascending order 1842, 3187, 4577, 7814. We see that the least number is 1842. We first assume $x=12748$ and $y=1842$ then multiply them and find the prime factorization of their product $xy$. If all the prime factors in prime factorization occur even number of times, then the product is a perfect square otherwise, we move to the next least numbers 3187, 4577, 7814.

Complete step-by-step answer:
The prime factorization of a composite number $X$ with $n$ prime factors ${{p}_{1}},{{p}_{2}},...{{p}_{n}}$ is given by
A perfect square is a number whose square root is an integer. We can find the square root of the number of $X$ by if all the prime factors ${{p}_{1}},{{p}_{2}},...{{p}_{n}}$ occur even number of times in the factorization because the prime factorization of $\sqrt{X}$ will have prime factors ${{p}_{1}},{{p}_{2}},...{{p}_{n}}$ will be occurring half of the number of times they occurred for $X$.\[\]
We also know that all the prime factors of $X$ will lie under $\sqrt{X}$ which means ${{p}_{1}},{{p}_{2}},...{{p}_{n}} < \sqrt{X}$. So we can check the primality of a number by dividing $X$ by all the primes less than $\sqrt{X}$.\[\]
Let us assume $x=12748$ and find its prime factorization. We see that 12748 is an even number, it must be divisible by 2. So, it can be written as $12748=2\times 6374$. Again, 6374 is an even number, so it is also divisible by 2, i.e., $6374=2\times 3187$. Now we have to check whether 3187 is a prime or not. We have square $\sqrt{3187}=56.45$. So we divide 3187 by all the primes less than $\sqrt{3187}$ which are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47, 53. We find that none of these primes exactly divides 3187 and so 3187 is a prime. So, we can write the prime factorization 12748 as \[12748=2\times 2\times 3187\]
Let us find the factorization of the least number of given in the options 1842. We have,
\[1842=2\times 3\times 307\]
We need to check whether 307 is a prime or not. We have $\sqrt{307}=17.5 < 18$. We check the divisibility of 307 by 2, 3, 5, 7, 11, 13, and 17 and conclude 307 is a prime. Now let us assume $y=307$ and the prime factorization of the product of $xy=12748\times 1842$. We have,
\[12748\times 1842=2\times 2\times 2\times 3\times 307 \times 3187 \]
We see that none of the prime factors occur an even number of times. So the product is not a perfect square and option D is incorrect. Now we move to the next least number 3187 in option A. We have already established that 3187 is prime and it will not have any prime factorization. We take $y=3187$ and find the prime factorization of the product of $xy=12748\times 3187$. We have
\[12748\times 3187=2\times 2\times 3187\times 3187\]
We see that both the prime factors 2 and 3187 occur an even number of times. The square root of the product will be $2\times 3187=6374$ which an integer is.
So when we multiply 12748 and 3187 the product will be a perfect square. So the correct option is A.

Note: We need to note that square root is always a positive quantity, so we have rejected the negative values in the problem but when we solve equations like ${{x}^{2}}=a, a> 0$ we have to take both the values because we are finding the roots. A perfect cube is a number whose cube root is an integer. There are other primality tests like the Fermat primality test in higher classes. We can represent any prime number in the form of (6n+1) or (6n-1). So, to check if the number 3187 is prime, we can try to represent it in this form. So, we can either have (3188-1) or (3186+1), out of these, 3186 is divisible by 6. So, we can represent 3187 in the form (6n+1), hence it is a prime number.