1.  Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225

As we can see from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Ans:- HCF of 225 and 135 is 45.



(ii) 196 and 38220 

This questios 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as the divisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore,

HCF(196, 38220) = 196.

Ans:- HCF of 196 and 38220 is 196.


(iii) 867 and 255

In this question 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as the divisor and applying the division lemma method, we get,

225 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51,

therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.

Ans:-  HCF of 867 and 225 is 51.

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2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Using Euclid division algorithm, we know that a=bq+r, 0≤ r ≤b ------(1)

Let a be any positive integer and b=6.

Then, by Euclid’s algorithm, a=6q+r for some integer q≥0, and r=0,1,2,3,4,5 ,or 0≤r<6.

Therefore, a=6q or 6q+1 or 6q+2 or 6q+3 or 6q+4 or 6q+5

6q+0:6 is divisible by 2, so it is an even number.

6q+1:6 is divisible by 2, but 1 is not divisible by 2 so it is an odd number.

6q+2:6 is divisible by 2, and 2 is divisible by 2 so it is an even number.

6q+3:6 is divisible by 2, but 3 is not divisible by 2 so it is an odd number.

6q+4:6 is divisible by 2, and 4 is divisible by 2 so it is an even number.

6q+5:6 is divisible by 2, but 5 is not divisible by 2 so it is an odd number.

 Odd integer can be expressed in the form 6q+1 or 6q+3 or 6q+5

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Exercise 1.1) 3.An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

HCF (616,32) is the maximum number of columns in which they can march.

Step 1: First find which integer is larger.

616>32

Step 2: Then apply the Euclid's division algorithm to 616 and 32 to obtain

616=32×19+8

Repeat the above step until you will get remainder as zero.

Step 3: Now consider the divisor 32 and the remainder 8, and apply the division lemma to get

32=8×4+0

Since the remainder is zero, we cannot proceed further.

Step 4: Hence the divisor at the last process is 8

So, the H.C.F. of 616 and 32 is 8.

 8 is the maximum number of columns in which they can march.

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4.  Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Let 'a' be any positive integer.

On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.

so that ,

a = 3q + r , where r = 0 ,1 , 2

When, r = 0

∴ a = 3q

When, r = 1

∴ a = 3q + 1

When, r = 2

∴ a = 3q + 2

When , a = 3q

On squaring both the sides,

a2=9q2

a2=3x3q2

a2=3

Where m = 3q2

When, a = 3q + 1

On squaring both the sides ,

a2=3q+12

a2=9q2+ 2 x 3q x 1+12

a2=9q2+6q+1

a2=3(3q2+2q)+1a2=3m+1m=3q2+2q

When, a = 3q + 2

On squaring both the sides,

a2= (3q+2)2

a2=3q2+2× 3q × 2 + 22

a2=9q2 + 12q+4

a2=(9q2+ 12+q+3) + 1

a2=3(3q2+ 4q+1) + 1

a2=3m+1
where m = 3q2 + 4q + 1

T square of any positive integer is either of the form 3m or 3m+1.

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5. Use Euclids division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
 
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
 
Where m is an integer such that m =   
Case 2: When a = 3q + 1,

a3=(3q+1)3

a3=27q3+27q2+9q+1
a3=9(3q3+3q2+q)+1
a3=9m+1

Where m is an integer such that m = (3q3+3q2+q)
Case 3: When a = 3q + 2,

a3=(3q +2)3
a3=27q3+ 54q2+ 36q + 8
a3= 9(3q3+ 6q2+ 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3+ 6q2+ 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

 

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