Finding the Least Number to Achieve a Perfect Square

The problem asks us to perform two main steps. First, find the smallest number, let's call it 'x', that we need to subtract from 10424 to get a perfect square. Second, find the smallest number by which 'x' should be multiplied to become a perfect square itself.

Let's tackle the first step: finding 'x'.

Step 1: Find the Least Number 'x' to Subtract

To find the least number 'x' that, when subtracted from 10424, results in a perfect square, we need to find the largest perfect square number that is less than 10424.

We can approximate the square root of 10424 to find which integer's square is close to it.

Calculating the square root: $$ \sqrt{10424} \approx 102.098 $$

The largest integer whose square is less than 10424 is the integer part of this square root, which is 102.

Now, let's find the square of 102: $$ 102^2 = 102 \times 102 = 10404 $$

This is the largest perfect square less than 10424.

The number 'x' that needs to be subtracted from 10424 to get this perfect square (10404) is the difference between 10424 and 10404.

$$ x = 10424 - 10404 = 20 $$

So, the least number 'x' to be subtracted is 20. Subtracting 20 from 10424 gives 10404, which is $102^2$.

Step 2: Find the Least Number to Multiply 'x' (which is 20) by to Get a Perfect Square

Now we know x = 20. We need to find the smallest number by which 20 should be multiplied to make it a perfect square.

To do this, we use the prime factorization of 20.

• Divide 20 by the smallest prime number, 2: $20 \div 2 = 10$
• Divide 10 by 2: $10 \div 2 = 5$
• Divide 5 by the next prime number, 5: $5 \div 5 = 1$

The prime factorization of 20 is: $$ 20 = 2 \times 2 \times 5 = 2^2 \times 5^1 $$

A number is a perfect square if and only if, in its prime factorization, the exponents of all the prime factors are even numbers.

In the factorization of 20 ($2^2 \times 5^1$), the exponent of the prime factor 2 is 2 (which is even), but the exponent of the prime factor 5 is 1 (which is odd).

To make the exponent of 5 even, we need to multiply 20 by another factor of 5.

Multiplying 20 by 5 gives: $$ 20 \times 5 = (2^2 \times 5^1) \times 5^1 = 2^2 \times 5^{1+1} = 2^2 \times 5^2 $$

This new number is $2^2 \times 5^2 = (2 \times 5)^2 = 10^2 = 100$.

100 is a perfect square (it's $10^2$). The least number we multiplied 20 by to get a perfect square is 5.

Therefore, the least number by which x (which is 20) should be multiplied to get a perfect square is 5.

Conclusion

The least number 'x' subtracted from 10424 to give a perfect square is 20. The least number by which 20 should be multiplied to get a perfect square is 5.

Revision Table: Perfect Squares and Factors

Additional Information: Making a Number a Perfect Square

When you have a number that is not a perfect square and you want to find the least number to multiply it by to make it a perfect square, follow these steps:

• Find the prime factorization of the number.
• Look at the exponents of each prime factor.
• If an exponent is odd, you need to multiply the number by that prime factor one more time to make the exponent even (e.g., $p^3$ needs $p^1$ to become $p^4$).
• If an exponent is even, you don't need to multiply by that prime factor for this purpose.
• The least number to multiply by is the product of all prime factors that had odd exponents, each raised to the power of 1.

For instance, if a number's prime factorization is $2^3 \times 3^1 \times 5^4$, the exponents of 2 (3) and 3 (1) are odd, while the exponent of 5 (4) is even. To make it a perfect square, you need to multiply by $2^1 \times 3^1 = 6$. The original number multiplied by 6 would be $(2^3 \times 3^1 \times 5^4) \times (2^1 \times 3^1) = 2^4 \times 3^2 \times 5^4$, which is a perfect square.