A Square and A Cube

 HEGP101 — A Square and A Cube

Complete Questions and Answers

This guide answers the exercises, in-text questions, “Figure it Out” questions, and the final activities appearing in the attached Grade 8 chapter.

 

1. Locker puzzle and factors

Question 1. Does every number have an even number of factors?

Answer: No. Every factor normally has a partner, but a perfect square has one unpaired factor: its square root. For example, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The factor 6 is paired with itself, while all other factors occur in pairs. Therefore, 36 has an odd number of factors.

 

Question 2. Which numbers have an odd number of factors?

Answer: Exactly the perfect squares have an odd number of factors: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, and so on.

 

Question 3. Write the locker numbers that remain open after all 100 people take their turns.

Answer: A locker remains open when its number has an odd number of factors. Hence, the open lockers are the perfect squares from 1 to 100:

 

Open locker numbers
1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Question 4. Which are the first five lockers toggled exactly twice?

Answer: A locker is toggled once for each factor of its number. A prime number has exactly two factors, 1 and itself. Therefore, the first five such lockers are 2, 3, 5, 7, and 11. The code is 2–3–5–7–11.

 

2. Square numbers and their patterns

Question 5. Can a square have side length 3.5 units or 2.5 units?

Answer: Yes. The areas are:

 

[ (3.5)^2=12.25\text{ square units},\qquad (2.5)^2=6.25\text{ square units}. ]

 

These are squares geometrically, although they are not perfect squares because their side lengths are not natural numbers.

 

Question 6. Complete the table of the first 30 perfect squares.

Number	Square	Number	Square	Number	Square
1	1	11	121	21	441
2	4	12	144	22	484
3	9	13	169	23	529
4	16	14	196	24	576
5	25	15	225	25	625
6	36	16	256	26	676
7	49	17	289	27	729
8	64	18	324	28	784
9	81	19	361	29	841
10	100	20	400	30	900

Question 7. What can be said about the units digit of a perfect square?

Answer: The units digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. Therefore, a number ending in 2, 3, 7, or 8 cannot be a perfect square. However, ending in 0, 1, 4, 5, 6, or 9 does not by itself prove that a number is a square.

 

Question 8. Write five numbers that are definitely not squares by looking at their units digits.

Answer: Examples include 12, 23, 37, 48, and 102. Each ends in 2, 3, 7, or 8.

 

Question 9. The squares 1², 9², 11², 19², 21², and 29² end in 1. Write the next two such squares.

Answer: The next two are:

 

[ 31^2=961,\qquad 39^2=1521. ]

 

Question 10. Which of the following have 6 in the units place: 38², 34², 46², 56², 74², 82²?

Answer: A square ends in 6 when its root ends in 4 or 6. Therefore:

 

[ 34^2=1156,\quad 46^2=2116,\quad 56^2=3136,\quad 74^2=5476. ]

 

So the answers are 34², 46², 56², and 74².

 

Question 11. If a number has three zeros at the end, how many zeros will its square have?

Answer: Its square will have six zeros at the end. For example, if the number is (a\times10^3), its square is (a^2\times10^6), provided (a) does not itself end in zero.

 

Question 12. What is the relationship between the trailing zeros of a number and those of its square? Can a square have an odd number of trailing zeros?

Answer: If a number has (n) trailing zeros, its square has (2n) trailing zeros. Therefore, a perfect square has an even number of trailing zeros: 0, 2, 4, 6, and so on.

 

Question 13. What is the relationship between the parity of a number and its square?

Answer: The square of an even number is even, and the square of an odd number is odd. In symbols, even² is even and odd² is odd.

 

Question 14. What are the differences between consecutive squares?

Answer:

 

[ 2^2-1^2=3,\quad 3^2-2^2=5,\quad 4^2-3^2=7,\quad 5^2-4^2=9. ]

 

In general,

 

[ (n+1)^2-n^2=2n+1. ]

 

Thus, the differences are consecutive odd numbers.

 

Question 15. What is the sum of the first (n) odd numbers?

Answer: The sum of the first (n) odd numbers is (n^2):

 

[ 1+3+5+\cdots +(2n-1)=n^2. ]

 

For example, (1+3+5+7+9=25=5^2).

 

Question 16. Given (35^2=1225), find (36^2) using odd numbers.

Answer: The next odd number after the first 35 odd numbers is the 36th odd number:

 

[ 2(36)-1=71. ]

 

Therefore,

 

[ 36^2=1225+71=1296. ]

 

Question 17. How can successive subtraction of odd numbers test whether a number is a perfect square?

Answer: Subtract 1, 3, 5, 7, and so on. If the process reaches exactly zero, the original number is a perfect square. If the result becomes negative before reaching zero, it is not a perfect square. For example, 25 reaches zero after five subtractions, so (25=5^2). The number 38 becomes negative after the seventh subtraction, so it is not a perfect square.

 

Question 18. How many numbers lie between two consecutive perfect squares?

Answer: Between (n^2) and ((n+1)^2), the number of integers is

 

[ (n+1)^2-n^2-1=2n. ]

 

Thus, there are 2n numbers between them.

 

Question 19. How many perfect squares lie in each block from 1 to 1000, and what is the largest square below 1000?

Answer:

 

Interval	Perfect squares in the interval	Count
1–100	1² through 10²	10
101–200	11² through 14²	4
201–300	15² through 17²	3
301–400	18² through 20²	3
401–500	21² through 22²	2
501–600	23² through 24²	2
601–700	25² through 26²	2
701–800	27² through 28²	2
801–900	29² through 30²	2
901–1000	31² through 31²	1

The largest perfect square less than 1000 is 31² = 961.

 

Question 20. What is the relation between triangular numbers and squares?

Answer: Consecutive triangular numbers add to a square:

 

[ 1+3=4=2^2,\quad 3+6=9=3^2,\quad 6+10=16=4^2. ]

 

In general, (T_n+T_{n+1}=(n+1)^2), where (T_n=\frac{n(n+1)}2). The next example is (10+15=25=5^2).

 

Question 21. What is the side of a square whose area is 49 cm²?

Answer: The side is (\sqrt{49}=7) cm.

 

Question 22. What are the square roots of 64?

Answer: The two integer square roots are +8 and −8, because (8^2=(-8)^2=64). The principal square root is (\sqrt{64}=8).

 

Question 23. Is 324 a perfect square? Is 156 a perfect square?

Answer:

 

[ 324=2^2\times3^4=(2\times3^2)^2=18^2, ]

 

so 324 is a perfect square and its square root is 18.

 

[ 156=2^2\times3\times13. ]

 

The prime factors 3 and 13 do not occur in pairs, so 156 is not a perfect square.

 

Question 24. Determine whether 1156 and 2800 are perfect squares using prime factorisation.

Answer:

 

[ 1156=2^2\times17^2=(2\times17)^2=34^2. ]

 

Therefore, 1156 is a perfect square and (\sqrt{1156}=34).

 

[ 2800=2^4\times5^2\times7. ]

 

The factor 7 is unpaired, so 2800 is not a perfect square.

 

3. Figure it Out: square numbers

Question 1. Which numbers are not perfect squares: 2032, 2048, 1027, 1089?

Answer: 2032, 2048, and 1027 are not perfect squares. The number (1089=33^2), so it is a perfect square.

 

Question 2. Which of (64^2,108^2,292^2,36^2) have last digit 4?

Answer: A number ending in 8 or 2 has a square ending in 4. Therefore,

 

[ 108^2\text{ and }292^2 ]

 

have last digit 4. The answer is 108² and 292².

 

Question 3. Given (125^2=15625), find (126^2).

Answer:

 

[126^2=(125+1)^2=125^2+2(125)(1)+1^2
=15625+250+1=15625+251=15876.]

 

The correct option is (iv) (15625+251), and the value is 15876.

 

Question 4. Find the side of a square whose area is 441 m².

Answer:

 

[ \text{side}=\sqrt{441}=21\text{ m}. ]

 

Question 5. Find the smallest square divisible by 4, 9, and 10.

Answer:

 

[ \operatorname{LCM}(4,9,10)=2^2\times3^2\times5=180. ]

 

To make 180 a square, multiply by 5:

 

[ 180\times5=900=30^2. ]

 

Therefore, the smallest square is 900.

 

Question 6. Find the smallest number by which 9408 must be multiplied to make a perfect square. Find the square root of the product.

Answer:

 

[ 9408=2^6\times3\times7^2. ]

 

Only the factor 3 has an odd exponent, so multiply by 3:

 

[ 9408\times3=28224=2^6\times3^2\times7^2=168^2. ]

 

The required multiplier is 3, and the square root is 168.

 

Question 7. How many numbers lie between the squares of 16 and 17, and of 99 and 100?

Answer:

 

[ 17^2-16^2-1=2(16)=32, ]

 

so there are 32 numbers between them.

 

[ 100^2-99^2-1=2(99)=198, ]

 

so there are 198 numbers between them.

 

Question 8. Fill in the missing numbers.

Answer: The pattern is

 

[ 1^2+2^2+2^2=3^2, ] [ 2^2+3^2+6^2=7^2, ] [ 3^2+4^2+12^2=13^2, ] [ 4^2+5^2+20^2=21^2, ] [ 9^2+10^2+90^2=91^2. ]

 

Thus, the blanks are 21, 90, and 91.

 

Question 9. How many tiny squares are in the picture? Give the prime factorisation.

Answer: The picture has 9 rows and 9 columns of motifs, giving (9\times9=81) motifs. Each motif contains (6\times6=36) tiny squares. Therefore,

 

[ 81\times36=2916. ]

 

The prime factorisation is

 

[ 2916=81\times36=3^4\times(2^2\times3^2)=2^2\times3^6. ]

 

Therefore, there are 2916 tiny squares, and the prime factorisation is (2^2\times3^6).

 

4. Cubic numbers

Question 10. How many unit cubes make a cube of side 2 cm? How many make one of side 3 cm?

Answer:

 

[ 2^3=8, ]

 

so a cube of side 2 cm contains 8 unit cubes. Also,

 

[ 3^3=27, ]

 

so a cube of side 3 cm contains 27 unit cubes.

 

Question 11. Why are 1, 8, 27, … called perfect cubes? Is 9 a cube?

Answer: They are called perfect cubes because they are products of a natural number multiplied by itself three times:

 

[ 1=1^3,\quad8=2^3,\quad27=3^3. ]

 

The number 9 is not a perfect cube, since it lies between (2^3=8) and (3^3=27).

 

Question 12. How many unit cubes are in a cube with edge length 4 units?

Answer:

 

[ 4^3=4\times4\times4=64. ]

 

There are 64 unit cubes.

 

Question 13. Complete the table of cubes up to 20³.

Number	Cube	Number	Cube	Number	Cube
1	1	8	512	15	3375
2	8	9	729	16	4096
3	27	10	1000	17	4913
4	64	11	1331	18	5832
5	125	12	1728	19	6859
6	216	13	2197	20	8000
7	343	14	2744

Question 14. What are the possible units digits of perfect cubes?

Answer: The possible units digits are 0, 1, 3, 5, 7, 8, and 9. A perfect cube cannot end in 2, 4, or 6.

 

Question 15. How many one-digit, two-digit, and three-digit perfect cubes are there?

Answer:

 

Number of digits	Perfect cubes	Count
One digit	1, 8	2
Two digits	27, 64	2
Three digits	125, 216, 343, 512, 729	5

Question 16. Can a cube end with exactly two zeros?

Answer: No. If a cube ends in zeros, the number of trailing zeros must be a multiple of 3, because ((10^k)^3=10^{3k}). Thus, a cube can end in 3, 6, 9, … zeros, but not exactly 2 zeros.

 

Question 17. Find the two representations of 4104 and 13832 as sums of two positive cubes.

Answer:

 

[ 4104=2^3+16^3=8+4096, ] [ 4104=9^3+15^3=729+3375. ]

 

Similarly,

 

[ 13832=2^3+24^3=8+13824, ] [ 13832=18^3+20^3=5832+8000. ]

 

Question 18. Find the sum (91+93+95+97+99+101+103+105+107+109) without adding each term separately.

Answer: There are 10 consecutive odd numbers. Their middle average is ((91+109)/2=100). Therefore,

 

[ 10\times100=1000. ]

 

The sum is 1000 = 10³.

 

Question 19. Find the cube roots of 64, 512, and 729.

Answer:

 

[ \sqrt[3]{64}=4,\qquad \sqrt[3]{512}=8,\qquad \sqrt[3]{729}=9. ]

 

Question 20. What happens to successive differences of perfect cubes?

Answer: For (1,8,27,64,125,216), the successive differences are:

 

Level	Differences
Cubes	1, 8, 27, 64, 125, 216
First differences	7, 19, 37, 61, 91
Second differences	12, 18, 24, 30
Third differences	6, 6, 6

The third differences are constant and equal to 6.

 

5. Figure it Out: cube numbers

Question 1. Find the cube roots of 27000 and 10648.

Answer:

 

[ 27000=30^3, ]

 

so (\sqrt[3]{27000}=30). Also,

 

[ 10648=22^3, ]

 

so (\sqrt[3]{10648}=22).

 

Question 2. What number should be multiplied by 1323 to make a cube number?

Answer:

 

[ 1323=3^3\times7^2. ]

 

The exponent of 7 must be raised from 2 to 3, so multiply by 7:

 

[ 1323\times7=9261=21^3. ]

 

The required number is 7.

 

Question 3. State true or false and explain.

Statement	Answer	Explanation
(i) The cube of any odd number is even.	False	An odd number cubed remains odd; for example, (3^3=27).
(ii) There is no perfect cube that ends with 8.	False	(2^3=8), and (12^3=1728).
(iii) The cube of a two-digit number may be a three-digit number.	False	The smallest two-digit number is 10, and (10^3=1000), which has four digits.
(iv) The cube of a two-digit number may have seven or more digits.	False	The largest is (99^3=970299), which has six digits.
(v) Cube numbers have an odd number of factors.	False	For example, 8 has four factors: 1, 2, 4, and 8.

Question 4. Find the cube roots of 1331, 4913, 12167, and 32768 without factorisation.

Answer:

 

[ 1331=11^3\Rightarrow\sqrt[3]{1331}=11, ] [ 4913=17^3\Rightarrow\sqrt[3]{4913}=17, ] [ 12167=23^3\Rightarrow\sqrt[3]{12167}=23, ] [ 32768=32^3\Rightarrow\sqrt[3]{32768}=32. ]

 

Question 5. Which is greatest: (67^3-66^3), (43^3-42^3), (67^2-66^2), or (43^2-42^2)?

Answer:

 

[ 67^3-66^3=13267, ] [ 43^3-42^3=5419, ] [ 67^2-66^2=133, ] [ 43^2-42^2=85. ]

 

Therefore, the greatest is (67^3-66^3).

 

6. Final activity: Square Pairs

Question 21. Arrange 1 to 17 in a row so that every adjacent pair adds to a square. Can it be done in more than one way?

Answer: Yes. One valid arrangement is:

 

[ 16,9,7,2,14,11,5,4,12,13,3,6,10,15,1,8,17. ]

 

The adjacent sums are 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, and 25—all perfect squares.

 

There are two solutions if reverse order is counted as a separate arrangement. The reverse is:

 

[ 17,8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16. ]

 

Apart from reversal, the arrangement is unique.

 

Question 22. Arrange 1 to 32 in a circle so that every adjacent pair adds to a square.

Answer: One valid circular arrangement is:

 

[1,8,28,21,4,32,17,19,30,6,3,13,12,24,25,11,
5,31,18,7,29,20,16,9,27,22,14,2,23,26,10,15.]

 

The last number also joins the first number, and (15+1=16), so the circular condition is satisfied. Reversing the direction gives the same circular arrangement in reverse order.

 

Key formulas used

Concept	Formula
Difference of consecutive squares	((n+1)^2-n^2=2n+1)
Sum of first (n) odd numbers	(1+3+\cdots +(2n-1)=n^2)
Numbers between consecutive squares	((n+1)^2-n^2-1=2n)
Cube of a number	(n^3=n\times n\times n)
Third finite difference of (n^3)	6

All numerical answers above have been checked against the exercise statements and the diagrams in the attached PDF.