Squares and Square Roots

The unit digit of a square is determined by the last digit of the number.
(i) 81: Ends in 1. 1² = 1. Unit digit is 1.
(ii) 272: Ends in 2. 2² = 4. Unit digit is 4.
(iii) 799: Ends in 9. 9² = 81. Unit digit is 1.
(iv) 3853: Ends in 3. 3² = 9. Unit digit is 9.
(v) 1234: Ends in 4. 4² = 16. Unit digit is 6.
(vi) 26387: Ends in 7. 7² = 49. Unit digit is 9.
(vii) 52698: Ends in 8. 8² = 64. Unit digit is 4.
(viii) 99880: Ends in 0. 0² = 0. Unit digit is 0.
(ix) 12796: Ends in 6. 6² = 36. Unit digit is 6.
(x) 55555: Ends in 5. 5² = 25. Unit digit is 5.

A natural number ending in 2, 3, 7, or 8, or having an odd number of zeroes at the end, is never a perfect square.
(i) 1057: Ends in 7.
(ii) 23453: Ends in 3.
(iii) 7928: Ends in 8.
(iv) 222222: Ends in 2.
(v) 64000: Ends in 3 zeroes (an odd number of zeroes).
(vi) 89722: Ends in 2.
(vii) 222000: Ends in 3 zeroes.
(viii) 505050: Ends in 1 zero.

The square of an odd number is an odd number, and the square of an even number is an even number.
(i) 431: It is an odd number. Hence, its square will be an odd number.
(ii) 2826: It is an even number. Its square will be an even number.
(iii) 7779: It is an odd number. Hence, its square will be an odd number.
(iv) 82004: It is an even number. Its square will be an even number.

Pattern: The number of zeroes between 1, 2, and 1 equals the number of zeroes in the middle of the original number.
100001² = 10000200001 (Four zeroes).
10000001² = 100000020000001 (Six zeroes).

Pattern: The digit in the middle reaches the number of 1's in the base number, counting up with a 0 in between.
For 1010101, there are four 1's, so it goes up to 4:
1010101² = 1020304030201
For 10203040504030201, it goes up to 5, so there should be five 1's:
101010101² = 10203040504030201

Pattern: n² + (n+1)² + [n(n+1)]² = [n(n+1) + 1]²
For the 4th line: 4² + 5² + (4×5)² = (20+1)² ⇒ 4² + 5² + 20² = 21²
For the 5th line: 5² + 6² + (5×6)² = (30+1)² ⇒ 5² + 6² + 30² = 31²
For the 6th line: 6² + 7² + (6×7)² = (42+1)² ⇒ 6² + 7² + 42² = 43².

Sum of first 'n' odd natural numbers = n².
(i) It has 5 odd numbers. Sum = 5² = 25.
(ii) It has 10 odd numbers. Sum = 10² = 100.
(iii) It has 12 odd numbers. Sum = 12² = 144.

(i) 49 = 7². So, 49 can be written as the sum of the first 7 odd numbers:
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13.

(ii) 121 = 11². So, 121 can be written as the sum of the first 11 odd numbers:
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.

The number of non-perfect square numbers lying between the squares of n and (n + 1) is 2n.
(i) For 12 and 13: Here n = 12. Numbers = 2 × 12 = 24.
(ii) For 25 and 26: Here n = 25. Numbers = 2 × 25 = 50.
(iii) For 99 and 100: Here n = 99. Numbers = 2 × 99 = 198.

Using the identity (a + b)² = a² + 2ab + b².
(i) 32: (30 + 2)² = 30² + 2(30)(2) + 2² = 900 + 120 + 4 = 1024.
(ii) 35: For numbers ending in 5, (a5)² = a(a+1) hundred + 25. Here a=3.
3(3+1) hundred + 25 = 3(4) hundred + 25 = 1200 + 25 = 1225.
(iii) 86: (80 + 6)² = 80² + 2(80)(6) + 6² = 6400 + 960 + 36 = 7396.
(iv) 93: (90 + 3)² = 90² + 2(90)(3) + 3² = 8100 + 540 + 9 = 8649.
(v) 71: (70 + 1)² = 70² + 2(70)(1) + 1² = 4900 + 140 + 1 = 5041.
(vi) 46: (40 + 6)² = 40² + 2(40)(6) + 6² = 1600 + 480 + 36 = 2116.

The form of a Pythagorean triplet is (2m, m² - 1, m² + 1).
(i) Let 2m = 6 ⇒ m = 3.
m² - 1 = 3² - 1 = 9 - 1 = 8.
m² + 1 = 3² + 1 = 9 + 1 = 10.
Triplet is (6, 8, 10).

(ii) Let 2m = 14 ⇒ m = 7.
m² - 1 = 7² - 1 = 49 - 1 = 48.
m² + 1 = 7² + 1 = 49 + 1 = 50.
Triplet is (14, 48, 50).

(iii) Let 2m = 16 ⇒ m = 8.
m² - 1 = 8² - 1 = 64 - 1 = 63.
m² + 1 = 8² + 1 = 64 + 1 = 65.
Triplet is (16, 63, 65).

(iv) Let 2m = 18 ⇒ m = 9.
m² - 1 = 9² - 1 = 81 - 1 = 80.
m² + 1 = 9² + 1 = 81 + 1 = 82.
Triplet is (18, 80, 82).

The unit digit of the square relates to the unit digit of its root.
(i) 9801: Number ends in 1. Possible unit digits of root are 1 or 9.
(ii) 99856: Number ends in 6. Possible unit digits of root are 4 or 6.
(iii) 998001: Number ends in 1. Possible unit digits of root are 1 or 9.
(iv) 657666025: Number ends in 5. Possible unit digit of root is 5.

A natural number ending in 2, 3, 7, or 8 is never a perfect square.
Hence, 153 (ends in 3), 257 (ends in 7), and 408 (ends in 8) are surely not perfect squares.
441 ends in 1, so it could be a perfect square.

(i) For 100:
100 - 1 = 99; 99 - 3 = 96; 96 - 5 = 91; 91 - 7 = 84; 84 - 9 = 75;
75 - 11 = 64; 64 - 13 = 51; 51 - 15 = 36; 36 - 17 = 19; 19 - 19 = 0.
Steps taken = 10. Therefore, √100 = 10.

(ii) For 169:
(Continuing similar subtraction of odd numbers from 1 to 25)
169 − 1 = 168, 168 − 3 = 165 ... 25 − 25 = 0.
Steps taken = 13. Therefore, √169 = 13.

(i) 729: 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3×3 × 3×3 × 3×3. √729 = 3 × 3 × 3 = 27.
(ii) 400: 400 = 2×2 × 2×2 × 5×5. √400 = 2 × 2 × 5 = 20.
(iii) 1764: 1764 = 2×2 × 3×3 × 7×7. √1764 = 2 × 3 × 7 = 42.
(iv) 4096: 4096 = 2¹². √4096 = 2⁶ = 64.
(v) 7744: 7744 = 2×2 × 2×2 × 2×2 × 11×11. √7744 = 2 × 2 × 2 × 11 = 88.
(vi) 9604: 9604 = 2×2 × 7×7 × 7×7. √9604 = 2 × 7 × 7 = 98.
(vii) 5929: 5929 = 7×7 × 11×11. √5929 = 7 × 11 = 77.
(viii) 9216: 9216 = 2¹° × 3². √9216 = 2⁵ × 3 = 32 × 3 = 96.
(ix) 529: 529 = 23×23. √529 = 23.
(x) 8100: 8100 = 2×2 × 3×3 × 3×3 × 5×5. √8100 = 2 × 3 × 3 × 5 = 90.

(i) 252: 252 = 2×2 × 3×3 × 7. Number is 7. New number = 1764. √1764 = 2 × 3 × 7 = 42.
(ii) 180: 180 = 2×2 × 3×3 × 5. Number is 5. New number = 900. √900 = 2 × 3 × 5 = 30.
(iii) 1008: 1008 = 2×2 × 2×2 × 3×3 × 7. Number is 7. New number = 7056. √7056 = 2 × 2 × 3 × 7 = 84.
(iv) 2028: 2028 = 2×2 × 3 × 13×13. Number is 3. New number = 6084. √6084 = 2 × 3 × 13 = 78.
(v) 1458: 1458 = 2 × 3×3 × 3×3 × 3×3. Number is 2. New number = 2916. √2916 = 2 × 3 × 3 × 3 = 54.
(vi) 768: 768 = 2⁸ × 3. Number is 3. New number = 2304. √2304 = 2⁴ × 3 = 16 × 3 = 48.

We divide by the unpaired prime factor.
(i) 252: 252 = 2² × 3² × 7. Divide by 7. New number = 36. √36 = 6.
(ii) 2925: 2925 = 3² × 5² × 13. Divide by 13. New number = 225. √225 = 3 × 5 = 15.
(iii) 396: 396 = 2² × 3² × 11. Divide by 11. New number = 36. √36 = 6.
(iv) 2645: 2645 = 5 × 23². Divide by 5. New number = 529. √529 = 23.
(v) 2800: 2800 = 2⁴ × 5² × 7. Divide by 7. New number = 400. √400 = 2² × 5 = 20.
(vi) 1620: 1620 = 2² × 3⁴ × 5. Divide by 5. New number = 324. √324 = 2 × 3² = 18.

Let number of students be x.
Therefore, money donated by each student = ₹ x.
Total money donated = x × x = x².
x² = 2401 ⇒ x = √2401.
Prime factorisation of 2401 = 7 × 7 × 7 × 7.
√2401 = 7 × 7 = 49.
Thus, number of students is 49.

Let number of rows = number of plants in each row = x.
Total number of plants = x × x = x².
x² = 2025 ⇒ x = √2025.
Prime factorisation of 2025 = 3 × 3 × 3 × 3 × 5 × 5.
√2025 = 3 × 3 × 5 = 45.
Number of rows = 45.

First find the LCM of 4, 9, 10.
LCM = 2 × 2 × 3 × 3 × 5 = 180.
Prime factors of 180 = 2×2 × 3×3 × 5.
The prime factor 5 is not in a pair. So, multiply 180 by 5 to make it a perfect square.
Smallest square number = 180 × 5 = 900.

LCM of 8, 15, 20 = 120.
Prime factors of 120 = 2×2 × 2 × 3 × 5.
Factors 2, 3, and 5 do not form pairs. Multiply 120 by (2 × 3 × 5) = 30.
Smallest square number = 120 × 30 = 3600.

Using long division method:
(i) 2304: 23 04. Largest square ≤ 23 is 16 (4²). Remainder = 7, bring down 04 = 704. New divisor 8_, 88 × 8 = 704. √2304 = 48.
(ii) 4489: 44 89. Largest square ≤ 44 is 36 (6²). Remainder = 8, bring down 89 = 889. New divisor 12_, 127 × 7 = 889. √4489 = 67.
(iii) 3481: 34 81. 5² = 25 ≤ 34. Remainder=9, bring down 81 = 981. Divisor 10_, 109 × 9 = 981. √3481 = 59.
(iv) 529: 5 29. 2² = 4 ≤ 5. Remainder=1, bring down 29 = 129. Divisor 4_, 43 × 3 = 129. √529 = 23.
(v) 3249: 32 49. 5² = 25 ≤ 32. Remainder=7, bring down 49 = 749. Divisor 10_, 107 × 7 = 749. √3249 = 57.

If a perfect square is of 'n' digits, then its square root will have (n/2) digits if n is even or ((n+1)/2) digits if n is odd.
(i) 64: n=2 (even). Digits in root = 2/2 = 1.
(ii) 144: n=3 (odd). Digits in root = (3+1)/2 = 2.
(iii) 4489: n=4 (even). Digits in root = 4/2 = 2.
(iv) 27225: n=5 (odd). Digits in root = (5+1)/2 = 3.
(v) 390625: n=6 (even). Digits in root = 6/2 = 3.

Using long division method:
(i) √2.56: √2 is 1 (remainder 1). Dividend matches 156 with divisor 26 (26×6=156). So, 1.6.
(ii) √7.29: √7 is 2 (remainder 3). Dividend matches 329 with divisor 47 (47×7=329). So, 2.7.
(iii) √51.84: √51 is 7 (remainder 2). Dividend matches 284 with divisor 142 (142×2=284). So, 7.2.
(iv) √42.25: √42 is 6 (remainder 6). Dividend matches 625 with divisor 125 (125×5=625). So, 6.5.
(v) √31.36: √31 is 5 (remainder 6). Dividend matches 636 with divisor 106 (106×6=636). So, 5.6.

Using long division, the remainder obtained is the number to be subtracted.
(i) 402: Quotient=20, Remainder=2. Subtract 2. Perfect square = 400. √400 = 20.
(ii) 1989: Quotient=44, Remainder=53. Subtract 53. Perfect square = 1936. √1936 = 44.
(iii) 3250: Quotient=57, Remainder=1. Subtract 1. Perfect square = 3249. √3249 = 57.
(iv) 825: Quotient=28, Remainder=41. Subtract 41. Perfect square = 784. √784 = 28.
(v) 4000: Quotient=63, Remainder=31. Subtract 31. Perfect square = 3969. √3969 = 63.

We find the square of the next quotient integer.
(i) 525: 22² = 484 < 525 < 23² = 529. To add: 529 - 525 = 4. Perfect square = 529, √529 = 23.
(ii) 1750: 41² = 1681 < 1750 < 42² = 1764. To add: 1764 - 1750 = 14. Perfect sq = 1764, √1764 = 42.
(iii) 252: 15² = 225 < 252 < 16² = 256. To add: 256 - 252 = 4. Perfect sq = 256, √256 = 16.
(iv) 1825: 42² = 1764 < 1825 < 43² = 1849. To add: 1849 - 1825 = 24. Perfect sq = 1849, √1849 = 43.
(v) 6412: 80² = 6400 < 6412 < 81² = 6561. To add: 6561 - 6412 = 149. Perfect sq = 6561, √6561 = 81.

Area of square = side × side = x².
x² = 441. So x = √441.
Using prime factorization or division method, √441 = 21.
Length of the side is 21 m.

By Pythagoras theorem, AC² = AB² + BC².
(a) AC² = 6² + 8² = 36 + 64 = 100 ⇒ AC = √100 = 10 cm.
(b) 13² = AB² + 5² ⇒ 169 = AB² + 25 ⇒ AB² = 169 - 25 = 144 ⇒ AB = √144 = 12 cm.

Let number of rows = number of columns = x. So total plants is x².
We check square root of 1000.
31² = 961 < 1000 < 32² = 1024.
To have equal rows and columns he needs to make the count up to a higher perfect square (1024).
Plants needed more = 1024 - 1000 = 24.

Total children = 500.
Since number of rows = number of columns, the arranged children will form a perfect square.
22² = 484 < 500 < 23² = 529.
The maximum number of children that can be arranged is 484 (forming a 22x22 grid).
Children left out = 500 - 484 = 16.

Subtract successive odd numbers from 225:
1) 225 - 1 = 224
2) 224 - 3 = 221
3) 221 - 5 = 216
4) 216 - 7 = 209
5) 209 - 9 = 200
6) 200 - 11 = 189
7) 189 - 13 = 176
8) 176 - 15 = 161
9) 161 - 17 = 144
10) 144 - 19 = 125
11) 125 - 21 = 104
12) 104 - 23 = 81
13) 81 - 25 = 56
14) 56 - 27 = 29
15) 29 - 29 = 0
Since we reach 0 after 15 steps, 225 is a perfect square, and √225 = 15.

Prime factorisation of 7056:
7056 = 2 × 3528 = 2 × 2 × 1764 = 2 × 2 × 2 × 882 = 2 × 2 × 2 × 2 × 441
441 = 3 × 147 = 3 × 3 × 49 = 3 × 3 × 7 × 7
So, 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Grouping in pairs and taking one from each pair:
√7056 = 2 × 2 × 3 × 7 = 84.

A natural number ending in 2, 3, 7, or 8 is never a perfect square.
(i) 153: Ends in 3. Not a perfect square.
(ii) 257: Ends in 7. Not a perfect square.
(iii) 408: Ends in 8. Not a perfect square.
(iv) 441: Ends in 1. It could be a perfect square (and is, 21²).

The unit digit of a square is determined by the square of the unit digit of the number.
(i) 272: Unit digit is 2. 2² = 4. So, unit digit of 272² is 4.
(ii) 799: Unit digit is 9. 9² = 81. So, unit digit is 1.
(iii) 3853: Unit digit is 3. 3² = 9. So, unit digit is 9.
(iv) 55555: Unit digit is 5. 5² = 25. So, unit digit is 5.

The number of non-perfect square numbers between n² and (n + 1)² is 2n.
Here, n = 45. Number of non-square numbers = 2 × 45 = 90.

The sum of the first 'n' odd natural numbers is n².
Since 49 = 7², it can be expressed as the sum of the first 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13.

For any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.
Let 2m = 8, then m = 4.
m² - 1 = 4² - 1 = 16 - 1 = 15.
m² + 1 = 4² + 1 = 16 + 1 = 17.
The Pythagorean triplet is (8, 15, 17).

First find the LCM of 8, 15, and 20.
LCM = 2 × 2 × 2 × 3 × 5 = 120.
Prime factorization of 120 = 2 × 2 × 2 × 3 × 5.
Here, factors 2, 3, and 5 do not form pairs.
To make it a perfect square, we must multiply by 2 × 3 × 5 = 30.
Smallest square number = 120 × 30 = 3600.

Prime factorisation of 2925:
2925 = 5 × 585 = 5 × 5 × 117 = 5 × 5 × 3 × 39 = 5 × 5 × 3 × 3 × 13
2925 = 3 × 3 × 5 × 5 × 13.
The prime factor 13 does not have a pair. Therefore, 2925 must be divided by 13 to make it a perfect square.
New perfect square = 2925 / 13 = 225.
Square root of new number = √225 = √(3 × 3 × 5 × 5) = 3 × 5 = 15.

Area of square = 5184 m².
Side of square = √5184 = 72 m.
Perimeter of square = 4 × side = 4 × 72 = 288 m.
Let breadth of rectangle be x m. Then length = 2x m.
Perimeter of rectangle = 2(length + breadth) = 2(2x + x) = 2(3x) = 6x m.
Given, 6x = 288 ⇒ x = 288 / 6 = 48 m.
Breadth = 48 m, Length = 2 × 48 = 96 m.
Area of rectangular field = length × breadth = 96 × 48 = 4608 m².

Place bars: 5 . 76
1. The largest square ≤ 5 is 4 (2²). Divisor = 2, Quotient = 2. Remainder = 5 - 4 = 1.
2. Bring down 76. Put a decimal in quotient. Dividend becomes 176.
3. Double the quotient (2 × 2 = 4). The new divisor is 4_.
4. Try 44 × 4 = 176. It matches.
So, quotient is 2.4. Therefore, √5.76 = 2.4.

Using long division to find √1300:
The nearest squares are 36² = 1296 and 37² = 1369.
We see that 36² < 1300 < 37².
To make 1300 a perfect square, we must add something to reach 37².
Number to be added = 37² - 1300 = 1369 - 1300 = 69.
The perfect square obtained is 1369. √1369 = 37.

Using long division on 5607:
- Pair from right: 56 07
- Nearest square to 56 is 49 (7²). Remainder = 7.
- Bring down 07. Dividend = 707. Double the quotient = 14.
- Try divisor 14_, 144 × 4 = 576. 145 × 5 = 725 (Too large).
- So we take 4. Remainder = 707 - 576 = 131.
Thus, 131 should be subtracted to get a perfect square.
Number to subtract = 131.
Perfect square = 5607 - 131 = 5476. √5476 = 74.

The greatest 4-digit number is 9999.
Find the square root of 9999 by long division.
Quotient comes out to be 99 with a remainder of 198.
Number we must subtract to make it a perfect square = 198.
Greatest 4-digit perfect square = 9999 - 198 = 9801. (√9801 = 99)

√12.25 = √(1225 / 100)
Prime factorisation of 1225 = 5 × 5 × 7 × 7.
√1225 = 5 × 7 = 35.
√100 = 10.
So, √12.25 = 35 / 10 = 3.5.

Using long division on 2.000000...
- Pair 0s: 2.00 00 00 00
- Quotient = 1. Remainder 1.
- Dividend 100, Divisor 2_, 24 × 4 = 96. Quotient 1.4, Rem 4.
- Dividend 400, Divisor 28_, 281 × 1 = 281. Quotient 1.41, Rem 119.
- Dividend 11900, Divisor 282_, 2824 × 4 = 11296. Quotient 1.414.
Thus, √2 ≈ 1.414.

By Pythagoras theorem, AC² = AB² + BC²
AC² = 12² + 5² = 144 + 25 = 169.
AC = √169 = 13 cm.

Total children = 500.
We need to find a perfect square near and less than 500.
Using long division on 500:
22² = 484. The remainder is 500 - 484 = 16.
So, 484 students can be arranged in 22 rows and 22 columns.
Children left out = 16.

The sum of the first n odd natural numbers is n².
Here, there are 8 consecutive odd numbers starting from 1 (n=8).
Sum = 8² = 64.

For any odd number n, n² can be expressed as the sum of (n² - 1)/2 and (n² + 1)/2.
Here n = 11. n² = 121.
First number = (121 - 1) / 2 = 120 / 2 = 60.
Second number = (121 + 1) / 2 = 122 / 2 = 61.
So, 121 = 60 + 61.