Question

Solve the given crossword puzzle and then fill up the given boxes.

Across

1. A natural number is called _____ if it is the square of some natural number.

3. There are _____ square numbers up to $200$ .

4. Find the square root of $1.21$ .

7. The square of $6000$ will have _____ zeroes.

Down

1. A perfect square can always be expressed as the product of pairs of _____.

2. The product of two square numbers is always a _____ number.

3. Find the smallest number by which $3645$ must be divided, so that it becomes a perfect square.

5. Find the smallest number by which $1152$ must be divided, so that it becomes a perfect square.

6. There is _____ square number between $50$ and $60$ .

Accepted Answer

Across

$1 .$ A natural number is called perfect square if it is the square of some natural number.

Thus, the answer is perfect square.

$3 .$ To compute the number of square numbers up to $200$ , we find $\sqrt{200}$ using the long division method.

$\begin{matrix} 1 & 4 \\ 1 & 2 & 00 \\ - & 1 \\ 24 & 1 & 00 \\ - & 96 \\ 4 \end{matrix}$

The remainder is $4$ . This shows that $14^{2}$ is less than $200$ by $4$ .

$\Rightarrow 14^{2} < 200$

Thus, the squares of all natural numbers till $14$ are less than $200$ .

Therefore, there are $14$ perfect squares less than $200$ .

Thus, the answer is fourteen.

$4 .$ To find the square root of $1.21$ , we use the long division method.

$\begin{matrix} 1 & . & 1 \\ 1 & 1 & . & 21 \\ - & 1 \\ 21 & 21 \\ - & 21 \\ 0 \end{matrix}$

Thus, $\sqrt{1.21} = 1.1$

Hence, the answer is one point one.

$7 .$ Consider the squares of some numbers, ending with different number of zeroes.

For example,

$50^{2} = 2500 300^{2} = 90000$

It may be observed that the square numbers have double the number of zeroes as present in the given number.

In general, if there are $n$ zeroes in each of the two (equal) numbers that are multiplied to obtain the square, the number of zeroes in the square will be $n + n = 2 n$ .

For $6000$ , $n = 3$ as $6000$ has three zeroes, and thus, $2 n = 6$

Thus, the number of zeroes in $6000^{2}$ will be six.

Hence, the answer is six.

Down

$1 .$ Any composite natural number can always be expressed as a product of prime factors. In the case of the square of a number, each prime factor occurs twice the number of times it occurs in the prime factorisation of the number itself. As the prime factors occur twice, they may be paired.

For exmaple,

$225 = 5 \times 45 = 5 \times 5 \times 9 = 5 \times 5 \times 3 \times 3 = 5^{2} \times 3^{2} = {(5 \times 3)}^{2} = 15^{2}$

Hence, a perfect square can always be expressed as the product of pairs of prime factors.

Thus, the answer is prime factors.

$2 .$ Consider the product of two square numbers, say $16 = 4^{2}$ and $25 = 5^{2}$ .

We can see that,

$16 \times 25 = 4^{2} \times 5^{2} = {(4 \times 5)}^{2} = 20^{2}$

Thus, the product of two square numbers is always a square number.

Hence, the answer is square.

$3 .$ We need to do the prime factorisation of $3645$ .

$3645 = 3 \times 1215 = 3 \times 3 \times 405 = 3 \times 3 \times 3 \times 135 = 3 \times 3 \times 3 \times 3 \times 45 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5$

By pairing the prime factors, we have

$3645 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 = {(3 \times 3 \times 3)}^{2} \times 5$

The prime factors need to be in pairs for a number to be a perfect square. In this case, as $3645$ has an unpaired factor $5$ , it is not a perfect square. We need to eliminate $5$ to make it a perfect square.

To get a factor of $3645$ , which is a perfect square, we divide $3645$ by $5$ , so that the resulting number will have all the prime factors in pairs.

$3645 \div 5 = {(3 \times 3 \times 3)}^{2}$ , which is a perfect square.

Thus, the answer is five.

$5 .$ We need to do the prime factorisation of $1152$ .

$1152 = 2 \times 576 = 2 \times 2 \times 288 = 2 \times 2 \times 2 \times 144 = 2 \times 2 \times 2 \times 12 \times 12 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2 \times 3$

By pairing the prime factors, we have

$1152 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = {(2 \times 2 \times 2 \times 3)}^{2} \times 2$

For a number to be a perfect square, the prime factors need to be in pairs. In this case, as $1152$ has an unpaired factor $2$ , it is not a perfect square. We need to eliminate $2$ to make it a perfect square.

To get a factor of $1152$ , which is a perfect square, we divide $1152$ by $2$ , so that the resulting number will have all the prime factors in pairs.

$1152 \div 2 = {(2 \times 2 \times 2 \times 3)}^{2}$ , which is a perfect square.

Thus, the answer is two.

$6 .$ We observe that $7^{2} = 49$ is a perfect square less than $50$ , and $8^{2} = 64$ , is a perfect square greater than $60$ . As $7$ and $8$ are consecutive natural numbers, there can be no perfect squares between $49$ and $64$ .

So, There are no square numbers between $50$ and $60$ .

Hence, the answer is no.

Therefore, the given crossword puzzle is,

Subject Experts Solutions for Chapter: Squares and Square Roots, Exercise 5: Activity

Subject Experts Mathematics Solutions for Exercise - Subject Experts Solutions for Chapter: Squares and Square Roots, Exercise 5: Activity

Questions from Subject Experts Solutions for Chapter: Squares and Square Roots, Exercise 5: Activity with Hints & Solutions

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