NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots (EX 6.3) Exercise 6.3

Q.1 What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Ans

(i) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) The one’s digit of the given number is 6. If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) The one’s digit of the given number is 5. If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

Q.2 Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Ans

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

(i) Since the number 153 has 3 in its unit’s place, so it is not a perfect square.

(ii) Since the number 257 has 7 in its unit’s place, so it is not a perfect square.

(iii) Since the number 408 has 8 in its unit’s place, so it is not a perfect square.

(iv) Since the number 441 has 1 in its unit’s place and a perfect square number may end with 1, so it is a perfect square.

Q.3 Find the square roots of 100 and 169 by the method of repeated subtraction.

Ans

\begin{array}{l}{\text{We know that the sum of the first n odd natural numbers is n}}^{\text{2}}\text{.}\\ \text{The square root of 169 can be obtained by the method of}\\ \text{repeated subtraction as follows}.\end{array}

\begin{array}{l}\text{Consider:}\\ \\ \left(\text{i}\right)\text{100}-\text{1}\text{= 99}\\ \left(\text{ii}\right)\text{99}-\text{3}\text{= 96}\\ \left(\text{iii}\right)\text{96}-\text{5}\text{= 91}\\ \left(\text{iv}\right)\text{91}-\text{7}\text{= 84}\\ \left(\text{v}\right)\text{84}-\text{9}\text{= 75}\\ \left(\text{vi}\right)\text{75}-\text{11}\text{= 64}\\ \left(\text{vii}\right)\text{64}-\text{13 = 51}\\ \left(\text{viii}\right)\text{51}-\text{15}\text{= 36}\\ \left(\text{ix}\right)\text{36}-\text{17}\text{= 19}\\ \left(\text{x}\right)\text{19}-\text{19}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting from}\\ {\text{1 to 100, and obtained 0 at 10}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{100}}\text{=10}\end{array}

$\begin{array}{l}\text{The square root of 169 can be obtained by the method}\\ \text{of repeated subtraction as follows:}\\ \\ \text{Consider:}\\ \\ \text{(i) 169}-\text{1}\text{= 168}\\ \left(\text{ii}\right)\text{168}-\text{3}\text{= 165}\\ \left(\text{iii}\right)\text{165}-\text{5}\text{= 160}\\ \left(\text{iv}\right)\text{160}-\text{7}\text{= 153}\\ \left(\text{v}\right)\text{153}-19\text{= 144}\\ \left(\text{vi}\right)\text{144}-\text{11}\text{= 133}\\ \left(\text{vii}\right)\text{133}-\text{13}\text{= 120}\\ \left(\text{viii}\right)\text{120}-\text{15 = 105}\\ \left(\text{ix}\right)\text{105}-\text{17}\text{= 88}\\ \left(\text{x}\right)\text{88}-\text{19}\text{= 69}\\ \left(\text{xi}\right)\text{69}-\text{21}\text{= 48}\\ \left(\text{xii}\right)\text{48}-\text{23}\text{= 25}\\ \left(\text{xiii}\right)\text{25}-\text{25}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting}\\ {\text{from 1 to 169, and obtained 0 at 13}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{169}}\text{=13}\end{array}$

Q.4 Find the square roots of the following numbers by the Prime Factorization Method.

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Ans

\begin{array}{l}\text{(i)}729\text{can be factorised as follows:}\\ \\ \begin{array}{cc}3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{729}=\underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{3}\times \text{3}}\\ \therefore \sqrt{729}=3\times 3\times 3=\text{27}\end{array}

\begin{array}{l}\text{(ii)400}\\ \\ \begin{array}{cc}2& 400\\ 2& 200\\ 2& 100\\ 5& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \text{400}=\underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{5}\times \text{5}}\\ \therefore \sqrt{400}=2\times 2\times 5=\text{20}\end{array}

\begin{array}{l}\text{(iii)1764}\\ \begin{array}{cc}2& 1764\\ 2& 882\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \text{1764}=\underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{7}\times \text{7}}\\ \therefore \sqrt{1764}=2\times 3\times 7=\text{42}\end{array}

\begin{array}{l}\text{(iv)}4096\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 4096\\ 2& 2048\\ 2& 1024\\ 2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ \text{4096}=\underset{¯}{2\times 2}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\\ \therefore \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2=\text{64}\\ \end{array}

\begin{array}{l}\text{(vi)}9604\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ \text{9604}=\underset{¯}{2\times 2}\times \underset{¯}{\text{7}\times \text{7}}\times \underset{¯}{\text{7}\times \text{7}}\\ \therefore \sqrt{9604}=2\times 7\times 7=\text{98}\\ \\ \text{(vii)5929}\\ \\ \begin{array}{cc}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}\\ \\ \text{5929}=\underset{¯}{7\times 7}\times \underset{¯}{\text{11}\times \text{11}}\\ \therefore \sqrt{5929}=7\times 11=\text{77}\end{array}

\begin{array}{l}\text{(viii) 9216}\\ \\ \begin{array}{cc}2& 9216\\ 2& 4608\\ 2& 2304\\ 2& 1152\\ 2& 576\\ 2& 288\\ 2& 144\\ 2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \text{9216}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{\text{3}\times 3}\\ \therefore \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3=\text{96}\\ \\ \text{(ix) 529}\\ \\ \begin{array}{cc}23& 529\\ 23& 23\\ & 1\end{array}\\ \\ \text{529}=\underset{¯}{23\times 23}\\ \therefore \sqrt{529}=\text{23}\end{array}

\begin{array}{l}\text{(x)}8100\\ \\ \begin{array}{cc}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 8100=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{8100}=2\times 3\times 3\times 5=90\\ \end{array}

Q.5 For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 252 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ 252\times 7=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}\\ \\ \therefore \text{252×7 = 1764 is a perfect square}\text{.}\\ \\ \therefore \sqrt{1764}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}=2\times 3\times 7=42\\ \end{array}$

$\begin{array}{l}\text{(ii)180 can be factorised as:}\\ \\ \begin{array}{cc}2& 180\\ 2& 90\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ 180=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we multiply the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 180 has to be multiplied with 5}\\ \text{to obtain a perfect square}\text{.}\\ \text{180}\times 5=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \text{180×5 = 900 is a perfect square}\text{.}\\ \therefore \sqrt{900}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}=2\times 3\times 5=30\\ \\ \text{(iii)1008 can be factorised as:}\\ \\ \begin{array}{cc}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\end{array}$

\begin{array}{l}1008=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 1008 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ \text{1008}\times 7=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}\\ \therefore \text{1008×7 = 7056 is a perfect square}\text{.}\\ \therefore \sqrt{7056}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}=2\times 2\times 3\times 7=84\\ \\ \text{(iv) 2028 can be factorised as follows:}\\ \begin{array}{cc}2& 2028\\ 2& 1014\\ 3& 507\\ 13& 169\\ 13& 13\\ & 1\end{array}\\ 2028=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2028 has to be multiplied with 3}\\ \text{to obtain a perfect square}\text{.}\\ \text{2028}\times 3=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times \underset{¯}{3\times 3}\\ \therefore \text{2028}\times 3\text{= 6084 is a perfect square}\text{.}\\ \therefore \sqrt{6084}=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times \underset{¯}{3\times 3}=2\times 3\times 13=78\end{array}

\begin{array}{l}\text{(v) 1458 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 1458=2\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \text{Here, the prime factor 2 does not have its pair}\text{.}\\ \text{If we multiply the given number by 2, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 1458 has to be multiplied with 2 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1458}\times 2=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \\ \therefore \text{1458}\times 2\text{= 2916 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2916}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}=2\times 3\times 3\times 3=54\\ \\ \end{array}

\begin{array}{l}\text{(vi) 768 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}\\ \\ 768=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 768 has to be multiplied with 3 to obtain}\\ \text{a perfect square}\text{.}\\ \text{768}\times 3=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \\ \therefore \text{768}\times 3\text{= 2304 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2304}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 2\times 2\times 2\times 3=48\end{array}

Q.6 For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 252 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{252}÷7=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{36}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 3=6\end{array}$

$\begin{array}{l}\text{(ii) 2925 can be factorised as follows:}\\ \\ \begin{array}{cc}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}\\ \\ 2925=\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\times 13\\ \text{Here, the prime factor 13 does not have its pair}\text{.}\\ \text{If we divide the given number by 13, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2925 has to be divided with 13 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2925}÷13=225\text{is a perfect square}\text{.}\\ \text{225 =}\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{225}=\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}=3\times 5=15\\ \\ \text{(iii) 396 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}\\ \end{array}$

$\begin{array}{l}396=\underset{¯}{3\times 3}\times \underset{¯}{2\times 2}\times 11\\ \text{Here, the prime factor 11 does not have its pair}\text{.}\\ \text{If we divide the given number by 11, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,396 has to be divided with 11 to obtain}\\ \text{a perfect square}\text{.}\\ \text{396}÷11=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{36}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 3=6\\ \\ \text{(iv) 2645 can be factorised as follows:}\\ \\ \begin{array}{cc}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}\\ \\ 2645=\underset{¯}{23\times 23}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,2645 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2645}÷5=529\text{is a perfect square}\text{.}\\ \text{529 =}\underset{¯}{23\times 23}\\ \therefore \sqrt{529}=23\end{array}$

$\begin{array}{l}\text{(v) 2800 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}\\ \\ 2800=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2800 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2800}÷7=400\text{is a perfect square}\text{.}\\ \text{400 =}\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{400}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}=2\times 2\times 5=20\end{array}$

$\begin{array}{l}\text{(vi) 1620 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ \\ 1620=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,1620 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1620}÷5=324\text{is a perfect square}\text{.}\\ \text{324 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{324}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}=2\times 3\times 3=18\end{array}$

Q.7 The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Ans

According to the question, each student donated as many rupees as the number of students of the class.
Therefore, the number of students in the class will be the square root of the amount donated by the students of the class.
The total amount of donation is Rs 2401.

\begin{array}{l}\text{Number of students in the class}=\sqrt{2400}\\ \\ \begin{array}{cc}7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ 2401=\underset{¯}{7\times 7}\times \underset{¯}{7\times 7}\\ \therefore \sqrt{2401}=7\times 7=49\end{array}

Hence, the number of students in the class is 49.

Q.8 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Ans

According to the question, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Also, total number of plants = 2025

Since, the number of rows and number of plants in each row are equal, therefore (Number of rows)² = 2025

$\begin{array}{l}\therefore \text{Number of rows =}\sqrt{2025}\\ \\ \begin{array}{cc}3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 2025=\underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{2025}=3\times 3\times 5=45\end{array}$

Hence, the number of rows and the number of plants in each row is 45.

Q.9 Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Ans

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{4},\text{9},\text{1}0\\ 2& \text{2},\text{9},\text{5}\\ 3& \text{1},\text{9},\text{5}\\ 3& \text{1},\text{3},\text{5}\\ 5& \text{1},\text{1},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900.

Q.10 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Ans

The number that will be perfectly divisible by each one of 8, 15, and 20 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{8},\text{15},\text{2}0\\ 2& \text{4},\text{15},\text{1}0\\ 2& \text{2},\text{15},\text{5}\\ 3& \text{1},\text{15},\text{5}\\ 5& \text{1},\text{5},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 8, 15, and 20 = 2 × 2 ×2 × 3 × 5 =120

Here, prime factors 2, 3 and 5 do not have their pairs. Therefore, 120 is not a perfect square. If we multiply 120 with 2×3×5, then the number will become a perfect square. Therefore, 120 should be multiplied with 2×3×5 to obtain a perfect square.

Hence, the required square number is 120×2×3×5 = 3600.