第4章多维数据集和多维数据集根–练习4.1 |套装1
问题12.以下数字必须除以最小的数字,以使商是一个理想的立方?
(i)675(ii)8640
(iii)1600(iv)8788
(v)7803(vi)107811
(vii)35721(viii)243000
解决方案:
(i) 675
Finding the prime factors of 675
675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
Hence, 675 is not a perfect cube.
We divide it by 52 = 25 to make the quotient a perfect cube, that gives 27 as quotient which is a perfect cube.
Hence, 25 is the required smallest number.
(ii) 8640
Finding the prime factors of 8640
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
= 23 × 23 × 33 × 5
Hence, 8640 is not a perfect cube.
We divide it by 5 to make the quotient a perfect cube, which gives 1728 as quotient which is a perfect cube.
Hence, 5 is the required smallest number.
(iii) 1600
Finding the prime factors of 1600
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 23 × 23 × 52
Hence, 1600 is not a perfect cube.
We divide it by 52 = 25 to make the quotient a perfect cube, which gives 64 as quotient which is a perfect cube
Hence, 25 is the required smallest number.
(iv) 8788
Finding the prime factors of 8788
8788 = 2 × 2 × 13 × 13 × 13
= 22 × 133
Hence, 8788 is not a perfect cube.
We divide it by 4 to make the quotient a perfect cube, which gives 2197 as quotient which is a perfect cube
Hence, 4 is the required smallest number.
(v) 7803
Finding the prime factors of 7803
7803 = 3 × 3 × 3 × 17 × 17
= 33 × 172
Hence, 7803 is not a perfect cube.
We divide it by 172 = 289 to make the quotient a perfect cube, which gives 27 as quotient which is a perfect cube.
Hence, 289 is the required smallest number.
(vi) 107811
Finding the prime factors of 107811
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
= 33 × 113 × 3
Hence, 107811 is not a perfect cube.
We divide it by 3 to make the quotient a perfect cube , which gives 35937 as quotient which is a perfect cube.
Hence, 3 is the required smallest number.
(vii) 35721
Finding the prime factors of 35721
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
= 33 × 33 × 72
Hence, 35721 is not a perfect cube.
We divide it by 72 = 49 to make the quotient a perfect cube, which gives 729 as quotient which is a perfect cube.
Hence, 49 is the required smallest number.
(viii) 243000
Finding the prime factors of 243000
243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
= 23 × 33 × 53 × 32
Hence, 243000 is not a perfect cube.
We divide it by 32 = 9 to make the quotient a perfect cube, which gives 27000 as quotient which is a perfect cube
Hence, 9 is the required smallest number.
问题13:证明如果一个数字是三倍的,那么它的立方是给定数字的立方的27倍。
解决方案:
Suppose the number is a
Therefore, its cube is = a3
Trebling the number = 3 × a = 3a
Therefore, the cube of new number = (3a) 3 = 27a3
This implies, the new cube is 27 times the original cube.
Hence, proved.
问题14.如果将数字乘以该数字的立方会发生什么
(i)3?
(ii)4?
(iii)5?
解决方案:
(i) 3?
Suppose the number is a
Therefore, its cube is = a3
Now, when the number is multiplied by 3
The new number becomes = 3a
Hence, the cube of new number is = (3a) 3 = 27a3
This implies, number will become 27 times the cube of the number.
(ii) 4?
Suppose the number is a
Therefore, its cube is = a3
Now, when the number is multiplied by 4
The new number becomes = 4a
Hence, the cube of new number is = (4a) 3 = 64a3
This implies, number will become 64 times the cube of the number.
(iii) 5?
Suppose the number is a
Therefore, its cube is = a3
Now, when the number is multiplied by 5
The new number becomes = 5a
Hence, the cube of new number is = (5a) 3 = 125a3
This implies number will become 125 times the cube of the number.
问题15.找到一个立方体的体积,该立方体的一个面的面积为64m 2 。
解决方案:
It is given that the area of one face of the cube is = 64 m2
Suppose the length of edge of cube be ‘a’ metres
a2 = 64
a = √ 64
= 8m
Now, volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
Hence, Volume of a cube is 512m3
问题16:找到一个表面积为384m 2的立方体的体积。
解决方案:
It is given that the surface area of cube is = 384 m2
Suppose the length of each edge of cube be ‘a’ meters
6a2 = 384
a2 = 384/6
= 64
a = √64
= 8m
Now, volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
Hence, Volume of a cube is 512m3
问题17:评估以下内容:
(i){(5 2 + 12 2 ) 1/2 } 3
(ii){(6 2 + 8 2 ) 1/2 } 3
解决方案:
(i) {(52 + 122)1/2}3
From the 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
From the above equation we get,
{(36 + 64)1/2}3
{(100)1/2}3
{(102)1/2}3
(10)3
1000
问题18.写下以下每个数字的立方的单位数字:
31,109,388,4276,5922,77774,44447,125125125
解决方案:
31
We will find the cube of unit digit only to find unit digit of cube of a number
Unit digit of 31 is 1
Cube of 1 = 13 = 1
Hence, the unit digit of cube of 31 is always 1
109
We will find the cube of unit digit only to find unit digit of cube of a number
Unit digit of 109 is = 9
Cube of 9 = 93 = 729
Hence, the unit digit of cube of 109 is always 9
388
We will find the cube of unit digit only to find unit digit of cube of a number.
Unit digit of 388 is = 8
Cube of 8 = 83 = 512
Hence, the unit digit of cube of 388 is always 2
4276
We will find the cube of unit digit only to find unit digit of cube of a number.
Unit digit of 4276 is = 6
Cube of 6 = 63 = 216
Hence, the unit digit of cube of 4276 is always 6
5922
We will find the cube of unit digit only to find unit digit of cube of a number.
Unit digit of 5922 is = 2
Cube of 2 = 23 = 8
Hence, the unit digit of cube of 5922 is always 8
77774
We will find the cube of unit digit only to find unit digit of cube of a number
Unit digit of 77774 is = 4
Cube of 4 = 43 = 64
Hence, the unit digit of cube of 77774 is always 4
44447
We will find the cube of unit digit only to find unit digit of cube of a number.
Unit digit of 44447 is = 7
Cube of 7 = 73 = 343
Hence, the unit digit of cube of 44447 is always 3
125125125
We will find the cube of unit digit only to find unit digit of cube of a number.
Unit digit of 125125125 is = 5
Cube of 5 = 53 = 125
Hence, the unit digit of cube of 125125125 is always 5
问题19.通过列方法查找以下数字的多维数据集:
(i)35
(ii)56
(iii)72
解决方案:
(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 |
Hence, 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 |
Hence, the cube of 56 is 175616
(iii) 72
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 |
Hence, the cube of 72 is 373248
问题20.以下哪个数字不是理想的立方体?
(i)64
(ii)216
(iii)243
(iv)1728年
解决方案:
(i) 64
Finding the prime factors of 64
64 = 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23
= 43
Therefore, it’s a perfect cube.
(ii) 216
Finding the prime factors of 216
216 = 2 × 2 × 2 × 3 × 3 × 3
= 23 × 33
= 63
Therefore, it’s a perfect cube.
(iii) 243
Finding the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3
= 33 × 32
Therefore, it’s not a perfect cube.
(iv) 1728
Finding the prime factors of 1728
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 23 × 23 × 33
= 123
Therefore, it’s a perfect cube.
问题21.对于Q中的每个非完美多维数据集。第20号找到它必须达到的最小数
(a)乘以使乘积是一个完美的立方体。
(b)除以使商是一个完美的立方体。
解决方案:
In the previous question the only non-perfect cube was = 243
(a) Multiplied so that the product is a perfect cube.
Finding the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Therefore, we should multiply it by 3 to make it a perfect cube.
(b) Divided so that the quotient is a perfect cube.
Finding the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Therefore, we have to divide it by 9 to make it a perfect cube.
问题22.通过取三个不同的值,n值可验证以下陈述的真实性:
(i)如果n为偶数,则n 3也为偶数。
(ii)如果n为奇数,则n 3也为奇数。
(ii)如果n除以3时剩下余数1,那么n 3除以3时还剩下1。
(iv)如果自然数n的形式为3p + 2,则n 3也是相同类型的数。
解决方案:
(i) If n is even, then n3 is also even.
Consider three even natural numbers 2, 4, 6
Hence, cubes of 2, 4 and 6 are
23 = 8
43 = 64
63 = 216
Therefore, we can see that all cubes are even in nature.
Hence proved.
(ii) If n is odd, then n3 is also odd.
Consider three odd natural numbers 3, 5, 7
Hence, cubes of 3, 5 and 7 are
33 = 27
53 = 125
73 = 343
Therefore, we can see that all cubes are odd in nature.
Hence proved.
(iii) If n is divided by 3 leaves remainder of 1, then when n3 is divided by 3 also leaves 1 as remainder.
Consider 4, 7 and 10 as three natural numbers of the form (3n+1)
Hence, cube of 4, 7, 10 are
43 = 64
73 = 343
103 = 1000
We get 1 as remainder in each case if we divide these numbers by 3.
Hence proved.
(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.
Consider 5, 8 and 11 as three natural numbers of the form (3p+2)
Hence, cube of 5, 8 and 10 are
53 = 125
83 = 512
113 = 1331
Let’s write these cubes in form of (3p + 2)
125 = 3 × 41 + 2
512 = 3 × 170 + 2
1331 = 3 × 443 + 2
Hence proved.
问题23.对以下陈述写“真”(T)或“假”(F):
(i)392是一个完美的立方体。
(ii)8640不是一个完美的立方体。
(iii)任何立方体都不能以正两个零结尾。
(iv)没有以4结尾的完美立方体。
(v)对于整数a,a 3始终大于a 2 。
(vi)如果a和b是使得a 2 > b 2的整数,则a 3 > b 3 。
(vii)如果a除以b,则a 3除以b 3 。
(viii)如果2以9结尾,则3以7结尾。
(ix)如果a 2以偶数个零结尾,则a 3以25结尾。
(x)如果a 2以偶数个零结尾,则a 3以奇数个零结尾。
解决方案:
(i) 392 is a perfect cube.
Finding the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 23 × 72
Therefore, the statement is False.
(ii) 8640 is not a perfect cube.
Finding the prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 23 × 33 × 5
Therefore, the statement is True
(iii) No cube can end with exactly two zeros.
Statement is True.
As a perfect cube always have zeros in multiple of 3.
(iv) There is no perfect cube which ends in 4.
It is known that 64 is a perfect cube = 4 × 4 × 4 and ends with 4.
Therefore, the statement is False.
(v) For an integer a, a3 is always greater than a2.
Statement is False 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.
Statement is False.
Because, in the case of the 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.
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.
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.
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.
Statement is False.
Since, when a = 100
a2 = 1002 = 10000 and a3 = 1003 = 1000000 (a3 doesn’t end with odd number of zeros)