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Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient.
A. Smaller number :5 Quotient : 3036 Square root :54
B. Smaller number :3 Quotient :3136 square root:56
C. Smaller number :7 Quotient :3236 square root:57
D. Data insufficient

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint: First we will write the given number as a product of its prime factor , then we will divide the number by the lowest factor such that in the prime product of the quotient the number of each prime factor is even.

Complete step by step solution:
Let’s write 9408 as its product of prime factors
$9408=2\times 2\times 2\times 2\times 2\times 2\times 3\times 7\times 7$
So we can write 9408 as ${{2}^{6}}{{3}^{1}}{{7}^{2}}$
Now we can see power of 2 is 6, power 3 is 1 and power of 7 is 2. Power of 2 and 7 is even, but power of 3 is odd so if we divide 9408 by 3 we will get a square number
The quotient when we divide 9408 by 3 we will get 3136 and the square root of 3136 is equal to 56, so the smaller number is 3 , quotient is 3136 and the square root is 56.

So, the correct answer is “Option B”.

Note: While writing evaluating the prime factor of any number check the divisibility of prime numbers such as if the sum of digits in the number is divisible by 3 the number is divisible by 3. If the last digit of the number from right is 0 or 5 then the number is divisible by 5. Multiply the last digit from right by 5 then add to the remaining number if it is divisible by 7 then the number is divisible by 7.