Exercise 1.1

Question 1: Use Euclid’s division algorithm to find the HCF of:

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

 (i) 135 and 225

 Given integers are 135 and 225 clearly 225 > 135 applying Euclid’s division lemma on 135 and 225.

                 We get

                 225 = 135 × + 90……………… (i)

                 Here remainder  So we again apply EDL on divisor 135 and remainder 90

                 135 = 90 × 1 + 45..................(ii)

                 Here, remainder  , so we apply  Euclid’s division lemma on divisor 90 and remainder 45

                 90 = 45 × 2 + 0 ………………….. (iii)

                 From equation (iii), remainder = 0. So the divisor at this stage and remainder of previous stage

                 i.e. 45 is HCF ( 135, 225) = 45

(ii) 196 and 38220

(ii) 196 and 38220

                 Given positive integers are 196 and 38220 and 38220 > 196 so applying EDL,

                 we get

                38220 = 196 × 195 + 0 …………… (i)

                Remainder at this stage is zero. So, the divisor of this stage i.e 196 is HCF of 38220 and 196

                HCF ( 196 , 38220) = 196

           (iii) 867 and 255

               Given positive integers are 867 and 255 and 867 > 255 So, applying Euclid’s division algorithm

               We get

               867 = 255 × 3 + 102  ………………. (i)

               Here, remainder . So, we again apply Euclid’s division algorithm on division 255 and remainder 102 .

Question 2: Show that any positive odd integer is of the form, 6q + 1 or 6q + 3, or 6q + 5 , where q is some integer.
Solution 2: .       Let a be any positive integer and b = 6.Then, by Euclid's algorithm a = 6q + r, for some integer q≥0 and where 0≤r<6 the possible remainders are 0, 1, 2, 3, 4, 5 i.e, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5,where q is the quotient. If a = 6q or 6q + 2 or 6q + 4, then a is an even integer. Also, an integer can be either even or odd.Therefore, any odd integer is of the form 6a + 1 or 6q + 3 or 6q + 5, where q is some integer. 
Question 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?
Solution 3:  HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid’s algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 
The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.
Question 4: Use Euclid’s division lemma to show that the square of any positive integer is either of 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.]
Solution 4:  Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q +2. Now, we have to prove that the squre of each of these can be written in the form 3m or 3m +1.             Now, (3q)2=9q2=3(3q2)=3m, where m=3q2             (3q+1)2=9q2+6q+1              =3(3q2+2q)+1             = 3m + 1, where m=3q2+2q             and, (3q+2q)2=9q2+12q+4              =3(3q2+4q+1)+1              = 3m + 1, where m=3q2+4q+1             Hence, the result.
Q.5     Use Euclid's division lemma to show that cube of any positive integer is either of the form 9q, 9q + 1 or 9q + 8.Sol.      Let x be any positive integer, then it is of  the form 3m, 3m + 1 or 3m +2. Now, we have prove that the cube of each of these can be rewritten in the form9q,  9q + 1 or 9q + 8.            Now, (3m)3=27m3=9(3m3)            = 9q, where q=3m3            (3m+1)3=(3m)3+3(3m)2.1+3(3m).12+1             =27m3+27m2+9m+1            =9(3m3+3m2+m)+1            = 9q + 1, where q=3m3+3m2+m            and (3m+2)3=(3m)3+3(3m)2.2+3(3m).22+8             =27m3+54m2+36m+8            =9(3m3+6m2+4m)+8            = 9q + 8, where q=3m3+6m2+4m

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