第 10 类 NCERT 解决方案 - 第 1 章实数 - 练习 1.1

问题 1. 使用欧几里得除法算法求 HCF：

(i) 135 和 225 (ii) 196 和 38220 (iii) 867 和 255

解决方案：

i)135 and 225

We need to apply Euclid’s algo, for finding HCF, now First Step to take Divisor ,we always take Divisor as smaller than Dividend.
Divisor=135
Dividend=225
Quotient=225/135=1(quotient is an integer value)
Euclid’s Div. Algo: 225=135*1+90
Since 90!=0 that means We have to again apply division lemma for remainder:90,Now Remainder Will become Divisor and Divisor
will become Dividend.
Euclid’s Div. Algo: 135=90*1+45
Again we Need to apply Division lemma because Yet remainder is not equal to 0 (45!=0) and just similar to prev. ways we will get
Divisor =45 and Dividend=90
Euclid’s Div. Algo:90=45*2+0
Here Remainder=0 that means We need to stop here,
When Remainder = 0 Then HCF = Divisor => HCF(225,135)=HCF(135,90)
=HCF(90,45)
=45.

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ii)196 and 38220

Divisor=196
Dividend=38220
Quotient=38220/196=195
Euclid’s Div. Algo: 38220=196*195+0
Remainder=0,We don’t need to apply Division lemma further,
HCF(38220,196)=196

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iii) 867 and 255

Divisor=255
Dividend=867
Quotient=867/225=3
Euclid’s Div. algo: 867=255*3+102
Remainder = 102 (!=0) That means Again we need to apply Division lemma method, Divisor=102,Dividend=255,Quotient=255/102=2
Euclid’s Div. algo: 255=102*2+51
Again Remainder =51(!=0), We need to apply Division lemma method again, Divisor=51,Dividend=102,Quotient=102/51=2
Euclid’S Div. algo:102=51*2+0
Remainder =0,We need to stop here ,
HCF(867,255)=HCF(255,102)
= HCF(102,51)
= 51

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问题 2.证明任何正奇数的形式为 6q + 1，或 6q + 3，或 6q + 5，其中 q 是某个整数。

解决方案：

**We know that any odd no integer is not divisible by 2,

When we divide 6q by 2 then it is perfectly divisible by 2 and give result 3q without any remainder,=>6q is even number

let take any positive integer x and y=6
as Per Euclid’s Div algo x=6q+r, q>=0 & 0<=r<6
if r=0 => x=6q, we earlier concluded that 6q is divisible by 2 =>x=6q is positive even integer

if r=2, 4 =>x=6q+2, x=6q+4 also An positive even integer because 6q is even as well as 2,4 also even and even divided by even gives
even

if r=1
Dividing 6q+1 by 2:
Dividend=6q+1,Divisor=2
Quotient = (6q+1)/2 =3q
as per Euclid’s div algo 6q+1=3q*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+1 is odd number, As q>=0 that means 6q+1 is positive odd integer

if r=3
Dividing 6q+3 by 2:
Dividend=6q+3,Divisor=2
Quotient = (6q+3)/2 =3q+1
as per Euclid’s div algo 6q+3=(3q+1)*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+3 is odd number, As q>=0 that means 6q+3 is positive odd integer

if r=5
Dividing 6q+5 by 2:
Dividend=6q+5,Divisor=2
Quotient = (6q+5)/2 =3q+2
as per Euclid’s div algo 6q+5=(3q+2)*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+5 is odd number, As q>=0 that means 6q+5 is positive odd integer
Overall Conclusion : Any positive integer x can be in form 6q+1,6q+3 or 6q+5 if it is odd otherwise it is in form 6q, 6q+2, 6q+4

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问题 3. 一支 616 人的军队要在一支 32 人的军乐队后面游行。两组行进的列数相同。他们可以行进的最大列数是多少？

解决方案：

Given,
Total number of army contingent members=616
Total number of army band members=32

Given two groups are to march in same number of column then maximum number of columns will be Highest Common Factor between two
groups ie HCF(616,32)

Finding HCF(616,32)
———————
Divisor=32,Dividend=616,quotient=616/32=19
Euclid’s Div. Algo: 616=32*19+8
Remainder = 8(!=0), again Apply Division lemma

Divisor=8,Dividend=32,Quotient=32/8=4
Euclid’s Div. Algo: 32=8*4+0
Remainder = 0,We have to stop here
HCF(616,32)=8

Therefore, Maximum Number of Columns Will be 8 in which Both groups can march

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问题 4. 使用欧几里得除法引理证明任何正整数的平方对于某个整数 m 的形式为 3m 或 3m + 1。

解决方案：

Let x be any positive integer, And y=3,
As per Euclid’s Div algo. x=3q+r;q>=0 && 0 <= r < 3 => r=(0,1,2)

If r=0: x=3q
Squaring both sides: x²=3*3*q²
let assume m1=3q²,Here m1 will be any positive integer because q>=0.
=>x²=3m1
if r=1: x=3q+1
Squaring both sides:x²=(3q+1)²
=>x² =3(3q² + 2q)+1
let assume m2=3q²+2q
=> x²= 3m2 + 1 ,Here m2 will be any positive integer because q>=0.
if r=2 : x=3q+2
Squaring both sides:x² =(3q+2)²
=>x² =3(3q² + 4q+1)+1
let assume m3=3q²+4q+1
=> x² = 3m3 + 1 ,Here m3 will be any positive integer because q>=0.
Since m1,m2,m3 are positive integer ,therefore we can Conclude that x²=3m or 3m+1 ,Where m is some integer.
This proved that the square of any positive integer(x) is either of the form 3m or 3m + 1 for some integer m

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问题 5. 用欧几里得除法引理证明任意正整数的立方都是 9m、9m + 1 或 9m + 8 的形式。

解决方案：

Let x be any positive integer, and y=3
As per Euclid’s Div algo. x=3q+r; q>=0 && 0 <= r < 3 => r=(0,1,2)

if r=0: x=3q
Cubing both sides: x³ =9*3*q³
let assume m1=3q³ ,Here m1 will be any positive integer because q>=0.
=>x2=9m1
if r=1: x=3q+1
Cubing both sides: x³ =(3q+1)³
=>x³ =9(3q³+3q²+q)+1
Let assume m2=3q³+3q²+q
=> x³ = 9m2 + 1 ,Here m2 will be any positive integer because q>=0.
if r=2 : x=3q+2
Cubing both sides: x³ =(3q+2)³
=>x³ =9(3q³ + 6q²+4q)+8
let assume m3=3q³+6q²+4q
=> x³ = 9m3 + 8 ,Here m3 will be any positive integer because q>=0.
Since m1,m2,m3 are positive integer ,therefore we can Conclude that x³=9m,9m+1 or 3m+8, where m is some integer.
This proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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