Exercise 1.1  Use Euclid’s Division algorithm to find the HCF of (1) 900 and 270 (ii) 196 and 38220 (iii) 1651 and 2032 .  Solution : (i) 900 and 720 Euclid division lemma a = bq + r , q >0 and  0 $$\leq r < b . 900 = 270 $$\times$$3 + 90 270 = 90 $$\times$$ 3 + 0 . The remainder is “0”  , so that we  cannot proceed further we conclude that HCF of (900,270) = 90 .   (ii) 196 and 38220 38220 = 196 $$\times$$+ 195 + 0  , Remainder = 0 The remainder is “0”  , so that we  cannot proceed further we conclude that HCF of (196 ,38220) = 90 . (iii) 1651 and 2032 2031 = 1651$$ \times$$ 1 +381 1651 = 381 $$\times$$4 +127 381 =127$$\times$$3 + 0 The remainder is “0”  , so that we  cannot proceed further we conclude that HCF of (2032 , 1951) = 127.   Using Euclid Lemma to show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+ 5 , where ‘q’ is some integer ? Solution : Let ‘a’ be and  odd positive integer  we  apply the division algorithm with a and b = 6 . Since 0 \leq r < 5 , the possible remainders are 0,1,2 ,3,4 and 5 . Ie a can be 6q or 6q + 1 or 6q+ 2 or 6q+3 or 6q+4 or 6q + 5 , where ‘q’ is a quotient. However , since a is odd  ; a cannot be 6q ,6q+2 ,6q+ 4 (since they are divisible by 2 ) . Therefore any odd integer is of the form 6q+ 1 (or) , 6q +2  (or) , 6q+3.   Using Euclid’s division of Lemma to show that square of any positive integer  of  the form  3p ,3p+1 (or) 3p + 2 .   Solution : Consider ‘a’ be the square of  an integer . Apply Euclid division lemma with ‘a’ and  ‘b’ = 3 . 0\leq r < 3 , the possible remainders  are 0,1,2. A= (3p) (or) (3p+1) (or) 3p+2 . Any square number is  of the from 3p ,3p+1 or 3p+3 , where ‘p’ is quotient. Using Euclid division lemma  to show that cube of  any positive integer  is of  the form9m ,9m+1 (or) 9m + 8 .  Solution : Let ‘a’ be the cube of a positive integer . Apply Euclid  division  Lemma for ‘a’ and b = 9 , A = bm + r , where ‘m’ is the quotient. a can be of the form 9m , 9m + 1 , 9m + 2 , 9m+3 ,-------,9m+8 . However ‘a’ is a cube . A cannot be (9m + 2 ,9m + 3 ,9m +4 ,9m +5,9m + 6,9m+7 ) Therefore cube of any positive integer of the form   9 m ,9m + 1 or  9m + 8. Show that one and only one out of n , n+2 and n+4 is divisible by 3 , where ‘n’ is a positive integer . Solution : Given numbers n , n+2 and n+ 4 . Case 1 :  ‘n’ is divisible by  3 , then ‘n’ is of the form 3k.             N+ 2 is 3k + 2  ---- leaves  a remainder ‘2’. N + 4 is 3k +4  =(3k+3) + 1 leaves a remainder ‘1’ when divided  by ‘3’. Case (2) : n + 2 is divisible by 3 , then n+2 is of the form 3k . Now n = 3k – 2 leaves a remainder  1  when divided  by 3 , N + 4 = n + 2 +2 = 3k + 2 ,leaves a remainder 2  when divided by ‘3’. Case (iii) : n +4 is divisible  by 3 . N + 4 = 3 k N = 3k – 4 = (3k-3) – 1 leaves a remainder 2  when divided  by ‘3’. N + 2 = 3k -2 = 3(k-1)+1 Leaves a remainder ‘1’ when divided  by ‘3’ . Therefore in all the above cases only one out of n ,n+2,n+4 is divisible by 3 .

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