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UNIT-1, NUMBER SYSTEMS, Numbers are intellectual witnesses that belong only to mankind., , 1. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x., (Ans:x=22), Ans: Using Euclid’s Division Lemma, a= bq+r , o ≤ r < b, 963=657×1+306, 657=306×2+45, 306=45×6+36, 45=36×1+9, 36=9×4+0, ∴ HCF (657, 963) = 9, now 9 = 657x + 963× (-15), 657x=9+963×15, =9+14445, 657x=14454, x=14454/657, x =22, , 2. Express the GCD of 48 and 18 as a linear combination., , i.e., ∴, , A=bq+r, where o ≤ r < b, 48=18x2+12, 18=12x1+6, 12=6x2+0, ∴ HCF (18,48) = 6, now 6= 18-12x1, 6= 18-(48-18x2), 6= 18-48x1+18x2, 6= 18x3-48x1, 6= 18x3+48x(-1), 6= 18x +48y, x=3 , y=-1, , 6, , (Ans: Not unique)
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6= 18×3 +48×(-1), =18×3 +48×(-1) + 18×48-18×48, =18(3+48)+48(-1-18), =18×51+48×(-19), 6=18x+48y, ∴, , x = 51, y = -19, , Hence, x and y are not unique., 3. Prove that one of every three consecutive integers is divisible by 3., Ans:, n,n+1,n+2 be three consecutive positive integers, We know that n is of the form 3q, 3q +1, 3q + 2, So we have the following cases, Case – I, , when n = 3q, , In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3, Case - II When n = 3q + 1, Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not, divisible by 3, Case – III When n = 3q +2, Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not, divisible by 3, Hence one of n, n + 1 and n + 2 is divisible by 3, , 4. Find the largest possible positive integer that will divide 398, 436, and 542 leaving, remainder 7, 11, 15 respectively., (Ans: 17), Ans: The required number is the HCF of the numbers, Find the HCF of 391, 425 and 527 by Euclid’s algorithm, ∴ HCF (425, 391) = 17, Now we have to find the HCF of 17 and 527, 527 = 17 х 31 +0, , 7
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∴ HCF (17,527) = 17, ∴ HCF (391, 425 and 527) = 17, 5. Find the least number that is divisible by all numbers between 1 and 10 (both, inclusive)., (Ans:2520), Ans: The required number is the LCM of 1,2,3,4,5,6,7,8,9,10, ∴ LCM = 2 × 2 × 3 × 2 × 3 × 5 × 7 = 2520, 6. Show that 571 is a prime number., Ans: Let x=571⇒√x=√571, Now 571 lies between the perfect squares of (23)2 and (24)2, Prime numbers less than 24 are 2,3,5,7,11,13,17,19,23, Since 571 is not divisible by any of the above numbers, 571 is a prime number, , 7. If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y., (Ans:5, -2 (Not unique), Ans: Using Euclid’s algorithm, the HCF (30, 72), 72 = 30 × 2 + 12, 30 = 12 × 2 + 6, 12 = 6 × 2 + 0, HCF (30,72) = 6, 6=30-12×2, 6=30-(72-30×2)2, 6=30-2×72+30×4, 6=30×5+72×-2, ∴ x = 5, y = -2, Also 6 = 30 × 5 + 72 (-2) + 30 × 72 – 30 × 72, Solve it, to get, x = 77, y = -32, Hence, x and y are not unique, , 8
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8. Show that the product of 3 consecutive positive integers is divisible by 6., Ans: Proceed as in question sum no. 3, 9. Show that for odd positive integer to be a perfect square, it should be of the form, 8k +1., Let a=2m+1, Ans: Squaring both sides we get, a2 = 4m (m +1) + 1, ∴ product of two consecutive numbers is always even, m(m+1)=2k, a2=4(2k)+1, a2 = 8 k + 1, Hence proved, , 10. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36., (Ans:999720), Ans: LCM of 24, 15, 36, LCM = 3, , ×2×2×2×3×, , 5 = 360, , Now, the greatest six digit number is 999999, Divide 999999 by 360, ∴ Q = 2777 , R = 279, ∴ the required number = 999999 – 279 = 999720, , 11. If a and b are positive integers. Show that √2 always lies between, , a 2 − 2b 2, a a + 2b, or, <, b( a + b ), b a+b, ∴ to compare these two number,, , Ans: We do not know whether, , Let us comute, , a a + 2b, −, b a+b, => on simplifying , we get, , 9, , a 2 − 2b 2, b( a + b ), , a, a − 2b, and, b, a+b
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∴, , a a + 2b, a a + 2b, −, > 0 or −, <0, b a+b, b a+b, , now, , a a + 2b, −, >0, b a+b, , a 2 − 2b 2, > 0 solve it , we get , a > √2b, b( a + b ), , when a > √2b and, a a + 2b, <, ,, b a+b, , Thus, , We have to prove that, , a + 2b, a, <√2<, a+b, b, , Now a >√2 b⇒2a2+2b2>2b2+ a2+2b2, On simplifying we get, a + 2b, a+b, Also a>√2, a, ⇒ >√2, b, √2>, , Similarly we get √2, <, Hence, , a + 2b, a+b, , a, a + 2b, <√2<, b, a+b, , 12. Prove that ( n − 1 + n + 1 ) is irrational, for every n∈N, , Self Practice, , 10
