In Exercise 4.1 of RD Sharma Class 8 Maths Chapter 4 Cubes and Cube Roots, we will explain problems based on the cubes of natural numbers and perfect cube numbers along with their properties, finding the cube of a two-digit number by the column method. Students can refer to and also download RD Sharma Solutions Class 8 Maths from the links provided below. RD Sharma Solutions provides ample questions along with their solutions, shortcut techniques and detailed explanations designed by our expert faculty, which will help students score well in the annual examination.
RD Sharma Solutions for Class 8 Maths Exercise 4.1 Chapter 4 Cubes and Cube Roots
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 4.1 Chapter 4 Cubes and Cube Roots
1. Find the cubes of the following numbers:
(i) 7 (ii) 12
(iii) 16 (iv) 21
(v) 40 (vi) 55
(vii) 100 (viii) 302
(ix) 301
Solution:
(i) 7
Cube of 7 is
7 = 7× 7 × 7 = 343
(ii) 12
Cube of 12 is
12 = 12× 12× 12 = 1728
(iii) 16
Cube of 16 is
16 = 16× 16× 16 = 4096
(iv) 21
Cube of 21 is
21 = 21 × 21 × 21 = 9261
(v) 40
Cube of 40 is
40 = 40× 40× 40 = 64000
(vi) 55
Cube of 55 is
55 = 55× 55× 55 = 166375
(vii) 100
Cube of 100 is
100 = 100× 100× 100 = 1000000
(viii) 302
Cube of 302 is
302 = 302× 302× 302 = 27543608
(ix) 301
Cube of 301 is
301 = 301× 301× 301 = 27270901
2. Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solutions:
Firstly let us find the Cube of natural numbers up to 10
13 = 1 × 1 × 1 = 1
23 = 2 × 2 × 2 = 8
33 = 3 × 3 × 3 = 27
43 = 4 × 4 × 4 = 64
53 = 5 × 5 × 5 = 125
63 = 6 × 6 × 6 = 216
73 = 7 × 7 × 7 = 343
83 = 8 × 8 × 8 = 512
93 = 9 × 9 × 9 = 729
103 = 10 × 10 × 10 = 1000
∴ From the above results, we can say that
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
3. Observe the following pattern:
13 = 1
13 + 23 = (1+2)2
13 + 23 + 33 = (1+2+3)2
Write the next three rows and calculate the value of 13 + 23 + 33 +…+ 93 by the above pattern.
Solution:
According to the given pattern,
13 + 23 + 33 +…+ 93
13 + 23 + 33 +…+ n3 = (1+2+3+…+n) 2
So when n = 10
13 + 23 + 33 +…+ 93 + 103 = (1+2+3+…+10) 2
= (55)2 = 55×55 = 3025
4. Write the cubes of 5 natural numbers, which are multiples of 3, and verify the followings:
“The cube of a natural number which is a multiple of 3 is a multiple of 27.’
Solution:
We know that the first 5 natural numbers, which are multiple of 3, are 3, 6, 9, 12 and 15
So now, let us find the cube of 3, 6, 9, 12 and 15
33 = 3 × 3 × 3 = 27
63 = 6 × 6 × 6 = 216
93 = 9 × 9 × 9 = 729
123 = 12 × 12 × 12 = 1728
153 = 15 × 15 × 15 = 3375
We found that all the cubes are divisible by 27
∴ “The cube of a natural number which is a multiple of 3 is a multiple of 27.’
5. Write the cubes of 5 natural numbers which are of form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:
“The cube of a natural number of the form 3n+1 is a natural number of the same form, i.e. when divided by 3, it leaves the remainder 1.”
Solution:
We know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16
So now, let us find the cube of 4, 7, 10, 13 and 16
43 = 4 × 4 × 4 = 64
73 = 7 × 7 × 7 = 343
103 = 10 × 10 × 10 = 1000
133 = 13 × 13 × 13 = 2197
163 = 16 × 16 × 16 = 4096
We found that all these cubes, when divided by ‘3’, leave the remainder 1.
∴ the statement “The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’ is true.
6. Write the cubes 5 natural numbers of form 3n+2(i.e.5,8,11….) and verify the following:
“The cube of a natural number of the form 3n+2 is a natural number of the same form, i.e. when it is divided by 3, the remainder is 2.’
Solution:
We know that the first 5 natural numbers in the form (3n + 2) are 5, 8, 11, 14 and 17
So now, let us find the cubes of 5, 8, 11, 14 and 17
53 = 5 × 5 × 5 = 125
83 = 8 × 8 × 8 = 512
113 = 11 × 11 × 11 = 1331
143 = 14 × 14 × 14 = 2744
173 = 17 × 17 × 17 = 4913
We found that all these cubes, when divided by ‘3’, leave the remainder 2.
∴ the statement“The cube of a natural number of the form 3n+2 is a natural number of the same form, i.e. when it is divided by 3, the remainder is 2’ is true.
7. Write the cubes of 5 natural numbers of which are multiples of 7, and verify the following:
“The cube of a multiple of 7 is a multiple of 73.
Solution:
The first 5 natural numbers, which are multiple of 7, are 7, 14, 21, 28 and 35
So, the Cube of 7, 14, 21, 28 and 35
73 = 7 × 7 × 7 = 343
143 = 14 × 14 × 14 = 2744
213 = 21× 21× 21 = 9261
283 = 28 × 28 × 28 = 21952
353 = 35 × 35 × 35 = 42875
We found that all these cubes are multiples of 73(343) as well.
∴The statement“The cube of a multiple of 7 is a multiple of 73 is true.
8. Which of the following are perfect cubes?
(i) 64 (ii) 216
(iii) 243 (iv) 1000
(v) 1728 (vi) 3087
(vii) 4608 (viii) 106480
(ix) 166375 (x) 456533
Solution:
(i) 64
First, find the factors of 64
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26 = (22)3 = 43
Hence, it’s a perfect cube.
(ii) 216
First, find thefactors of 216
216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 = 63
Hence, it’s a perfect cube.
(iii) 243
First, find thefactors of 243
243 = 3 × 3 × 3 × 3 × 3 = 35 = 33 × 32
Hence, it’s not a perfect cube.
(iv) 1000
First, find thefactors of 1000
1000 = 2 × 2 × 2 × 5 × 5 × 5 = 23 × 53 = 103
Hence, it’s a perfect cube.
(v) 1728
First, find thefactors of 1728
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 26 × 33 = (4 × 3 )3 = 123
Hence, it’s a perfect cube.
(vi) 3087
First, find thefactors of 3087
3087 = 3 × 3 × 7 × 7 × 7 = 32 × 73
Hence, it’s not a perfect cube.
(vii) 4608
First, find thefactors of 4608
4608 = 2 × 2 × 3 × 113
Hence, it’s not a perfect cube.
(viii) 106480
First, find thefactors of 106480
106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11
Hence, it’s not a perfect cube.
(ix) 166375
First, find thefactors of 166375
166375= 5 × 5 × 5 × 11 × 11 × 11 = 53 × 113 = 553
Hence, it’s a perfect cube.
(x) 456533
First, find thefactors of 456533
456533= 11 × 11 × 11 × 7 × 7 × 7 = 113 × 73 = 773
Hence, it’s a perfect cube.
9. Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
Solution:
(i) 216 = 23 × 33 = 63
It’s a cube of even natural number.
(ii) 512 = 29 = (23)3 = 83
It’s a cube of even natural number.
(iii) 729 = 33 × 33 = 93
It’s not a cube of even natural number.
(iv) 1000 = 103
It’s a cube of even natural number.
(v) 3375 = 33 × 53 = 153
It’s not a cube of even natural number.
(vi) 13824 = 29 × 33 = (23)3 × 33 = 83×33 = 243
It’s a cube of even natural number.
10. Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
Solution:
(i) 125 = 5 × 5 × 5 × 5 = 53
It’s a cube of odd natural number.
(ii) 343 = 7 × 7 × 7 = 73
It’s a cube of odd natural number.
(iii) 1728 = 26 × 33 = 43 × 33 = 123
It’s not a cube of odd natural number. As 12 is an even number.
(iv) 4096 = 212 = (26)2 = 642
It’s not a cube of odd natural number. As 64 is an even number.
(v) 32768 = 215 = (25)3 = 323
It’s not a cube of odd natural number. As 32 is an even number.
(vi) 6859 = 19 × 19 × 19 = 193
It’s a cube of odd natural number.
11. What is the smallest number by which the following numbers must be multiplied so that the products are perfect cubes?
(i) 675 (ii) 1323
(iii) 2560 (iv) 7803
(v) 107811 (vi) 35721
Solution:
(i) 675
First, find the factors of 675
675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
∴To make a perfect cube, we need to multiply the product by 5.
(ii) 1323
First, find the factors of 1323
1323 = 3 × 3 × 3 × 7 × 7
= 33 × 72
∴To make a perfect cube, we need to multiply the product by 7.
(iii) 2560
First, find the factors of 2560
2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5
= 23 × 23 × 23 × 5
∴To make a perfect cube, we need to multiply the product by 5 × 5 = 25.
(iv) 7803
First, find the factors of 7803
7803 = 3 × 3 × 3 × 17 × 17
= 33 × 172
∴To make a perfect cube, we need to multiply the product by 17.
(v) 107811
First, find the factors of 107811
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
= 33 × 3 × 113
∴To make a perfect cube, we need to multiply the product by 3 × 3 = 9.
(vi) 35721
First, find the factors of 35721
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
= 33 × 33 × 72
∴ To make a perfect cube, we need to multiply the product by 7.
12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
(i) 675 (ii) 8640
(iii) 1600 (iv) 8788
(v) 7803 (vi) 107811
(vii) 35721 (viii) 243000
Solution:
(i) 675
First, find the prime factors of 675
675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
Since 675 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 52 = 25, which gives 27 as the quotient where, 27 is a perfect cube.
∴ 25 is the required smallest number.
(ii) 8640
First, find the prime factors of 8640
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
= 23 × 23 × 33 × 5
Since 8640 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 5, which gives 1728 as the quotient, and we know that 1728 is a perfect cube.
∴ 5 is the required smallest number.
(iii) 1600
First, find the prime factors of 1600
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 23 × 23 × 52
Since 1600 is not a perfect cube,
To make the quotient a perfect cube, we divide it by 52 = 25, which gives 64 as the quotient, and we know that 64 is a perfect cube
∴ 25 is the required smallest number.
(iv) 8788
First, find the prime factors of 8788
8788 = 2 × 2 × 13 × 13 × 13
= 22 × 133
Since 8788 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 4, which gives 2197 as the quotient, and we know that 2197 is a perfect cube
∴ 4 is the required smallest number.
(v) 7803
First, find the prime factors of 7803
7803 = 3 × 3 × 3 × 17 × 17
= 33 × 172
Since 7803 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 172 = 289, which gives 27 as the quotient, and we know that 27 is a perfect cube
∴ 289 is the required smallest number.
(vi) 107811
First, find the prime factors of 107811
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
= 33 × 113 × 3
Since 107811 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 3, which gives 35937 as the quotient, and we know that 35937 is a perfect cube.
∴ 3 is the required smallest number.
(vii) 35721
First, find the prime factors of 35721
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
= 33 × 33 × 72
Since 35721 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 72 = 49, which gives 729 as the quotient, and we know that 729 is a perfect cube
∴ 49 is the required smallest number.
(viii) 243000
First, find the prime factors of 243000
243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
= 23 × 33 × 53 × 32
Since 243000 is not a perfect cube.
To make the quotient a perfect cube, we divide it by 32 = 9, which gives 27000 as the quotient and we know that 27000 is a perfect cube
∴ 9 is the required smallest number.
13. Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let us consider a number as a
So the cube of the assumed number is = a3
Now, the number is trebled = 3 × a = 3a
So the cube of new number = (3a) 3 = 27a3
∴ The new cube is 27 times of the original cube.
Hence, proved.
14. What happens to the cube of a number if the number is multiplied by
(i) 3?
(ii) 4?
(iii) 5?
Solution:
(i) 3?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 3
New number becomes = 3a
So the cube of new number will be = (3a) 3 = 27a3
Hence, the number will become 27 times the cube of the number.
(ii) 4?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 4
New number becomes = 4a
So the cube of new number will be = (4a) 3 = 64a3
Hence, the number will become 64 times the cube of the number.
(iii) 5?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 5
New number becomes = 5a
So the cube of new number will be = (5a) 3 = 125a3
Hence, the number will become 125 times the cube of the number.
15. Find the volume of a cube, one face of which has an area of 64m2.
Solution:
We know that the given area of one face of the cube = 64 m2
Let the length of the edge of the cube be ‘a’ metre
a2 = 64
a = √ 64
= 8m
Now, the volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
∴ The volume of a cube is 512m3
16. Find the volume of a cube whose surface area is 384m2.
Solution:
We know that the surface area of the cube = 384 m2
Let us consider the length of each edge of the cube be ‘a’ meter
6a2 = 384
a2 = 384/6
= 64
a = √64
= 8m
Now, the volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
∴ The volume of a cube is 512m3
17. Evaluate the following:
(i) {(52 + 122)1/2}3
(ii) {(62 + 82)1/2}3
Solution:
(i) {(52 + 122)1/2}3
When simplified above equation we get,
{(25 + 144)1/2}3
{(169)1/2}3
{(132)1/2}3
(13)3
2197
(ii) {(62 + 82)1/2}3
When simplified above equation we get,
{(36 + 64)1/2}3
{(100)1/2}3
{(102)1/2}3
(10)3
1000
18. Write the units digit of the cube of each of the following numbers:
31, 109, 388, 4276, 5922, 77774, 44447, 125125125
Solution:
31
To find the unit digit of the cube of a number, we perform the cube of the unit digit only.
Unit digit of 31 is 1
Cube of 1 = 13 = 1
∴ The unit digit of the cube of 31 is always 1
109
To find the unit digit of the cube of a number, we perform the cube of unit digit only.
Unit digit of 109 is = 9
Cube of 9 = 93 = 729
∴ Unit digit of cube of 109 is always 9
388
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 388 is = 8
Cube of 8 = 83 = 512
∴ Unit digit of cube of 388 is always 2
4276
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 4276 is = 6
Cube of 6 = 63 = 216
∴ Unit digit of cube of 4276 is always 6
5922
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 5922 is = 2
Cube of 2 = 23 = 8
∴ Unit digit of cube of 5922 is always 8
77774
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 77774 is = 4
Cube of 4 = 43 = 64
∴ Unit digit of cube of 77774 is always 4
44447
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 44447 is = 7
Cube of 7 = 73 = 343
∴ Unit digit of cube of 44447 is always 3
125125125
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 125125125 is = 5
Cube of 5 = 53 = 125
∴ Unit digit of cube of 125125125 is always 5
19. Find the cubes of the following numbers by column method:
(i) 35
(ii) 56
(iii) 72
Solution:
(i) 35
We have, a = 3 and b = 5
| Column I
a3 |
Column II
3×a2×b |
Column III
3×a×b2 |
Column IV
b3 |
| 33 = 27 | 3×9×5 = 135 | 3×3×25 = 225 | 53 = 125 |
| +15 | +23 | +12 | 125 |
| 42 | 158 | 237 | |
| 42 | 8 | 7 | 5 |
∴ The cube of 35 is 42875
(ii) 56
We have, a = 5 and b = 6
| Column I
a3 |
Column II
3×a2×b |
Column III
3×a×b2 |
Column IV
b3 |
| 53 = 125 | 3×25×6 = 450 | 3×5×36 = 540 | 63 = 216 |
| +50 | +56 | +21 | 126 |
| 175 | 506 | 561 | |
| 175 | 6 | 1 | 6 |
∴ The cube of 56 is 175616
(iii) 72
We have, a = 7 and b = 2
| Column I
a3 |
Column II
3×a2×b |
Column III
3×a×b2 |
Column IV
b3 |
| 73 = 343 | 3×49×2 = 294 | 3×7×4 = 84 | 23 = 8 |
| +30 | +8 | +0 | 8 |
| 373 | 302 | 84 | |
| 373 | 2 | 4 | 8 |
∴ The cube of 72 is 373248
20. Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64
Firstly let us find the prime factors of 64
64 = 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23
= 43
Hence, it’s a perfect cube.
(ii) 216
Firstly let us find the prime factors of 216
216 = 2 × 2 × 2 × 3 × 3 × 3
= 23 × 33
= 63
Hence, it’s a perfect cube.
(iii) 243
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3
= 33 × 32
Hence, it’s not a perfect cube.
(iv) 1728
Firstly let us find the prime factors of 1728
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 23 × 23 × 33
= 123
Hence, it’s a perfect cube.
21. For each of the non-perfect cubes in Q. No 20, find the smallest number by which it must be
(a) Multiplied so that the product is a perfect cube.
(b) Divided so that the quotient is a perfect cube.
Solution:
The only non-perfect cube in the previous question was = 243
(a) Multiplied so that the product is a perfect cube.
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Hence, to make it a perfect cube, we should multiply it by 3.
(b) Divided so that the quotient is a perfect cube.
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Hence, to make it a perfect cube, we have to divide it by 9.
22. By taking three different values of n, verify the truth of the following statements:
(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(ii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as the remainder when divided by 3.
(iv) If a natural number n is of form 3p+2, then n3 is also a number of the same type.
Solution:
(i) If n is even, then n3 is also even.
Let us consider three even natural numbers 2, 4, 6
So now, Cubes of 2, 4 and 6 are
23 = 8
43 = 64
63 = 216
Hence, we can see that all cubes are even in nature.
The statement is verified.
(ii) If n is odd, then n3 is also odd.
Let us consider three odd natural numbers 3, 5, 7
So now, cubes of 3, 5 and 7 are
33 = 27
53 = 125
73 = 343
Hence, we can see that all cubes are odd in nature.
The statement is verified.
(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as the remainder when divided by 3.
Let us consider three natural numbers of the form (3n+1) are 4, 7 and 10
So now, cube of 4, 7, 10 are
43 = 64
73 = 343
103 = 1000
We can see that if we divide these numbers by 3, we get 1 as the remainder in each case.
Hence, the statement is verified.
(iv) If a natural number n is of form 3p+2, then n3 is also a number of the same type.
Let us consider three natural numbers of the form (3p+2) are 5, 8 and 11
So now, cube of 5, 8 and 10 are
53 = 125
83 = 512
113 = 1331
Now, we try to write these cubes in the form of (3p + 2)
125 = 3 × 41 + 2
512 = 3 × 170 + 2
1331 = 3 × 443 + 2
Hence, the statement is verified.
23. Write true (T) or false (F) for the following statements:
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a2.
(vi) If a and b are integers such that a2>b2, then a3>b3.
(vii) If a divides b, then a3 divides b3.
(viii) If a2 ends in 9, then a3 ends in 7.
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.
Solution:
(i) 392 is a perfect cube.
Firstly let’s find the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 23 × 72
Hence the statement is False.
(ii) 8640 is not a perfect cube.
Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 23 × 33 × 5
Hence the statement is True
(iii) No cube can end with exactly two zeros.
The statement is True.
Because a perfect cube always has zeros in multiple of 3.
(iv) There is no perfect cube which ends in 4.
We know 64 is a perfect cube = 4 × 4 × 4, and it ends with 4.
Hence the statement is False.
(v) For an integer a, a3 is always greater than a2.
The statement is False.
Because in the case of negative integers,
(-2)2 = 4 and (-2)3 = -8
(vi) If a and b are integers such that a2>b2, then a3>b3.
The statement is False.
In the case of negative integers,
(-5)2 > (-4)2 = 25 > 16
But, (-5)3 > (-4)3 = -125 > -64 is not true.
(vii) If a divides b, then a3 divides b3.
The statement is True.
If a divides b
b/a = k, so b=ak
b3/a3 = (ak)3/a3 = a3k3/a3 = k3,
For each value of b and a, its true.
(viii) If a2 ends in 9, then a3 ends in 7.
The statement is False.
Let a = 7
72 = 49 and 73 = 343
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
The statement is False.
Since when a = 20
a2 = 202 = 400 and a3 = 8000 (a3 doesn’t end with 25)
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.
The statement is false.
Since when a = 100
a2 = 1002 = 10000 and a3 = 1003 = 1000000 (a3 doesn’t end with odd number of zeros)
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