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Real number by G R Ahmed of KVK

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MD. G R Ahmed
MD. G R Ahmed

Here are the step-by-step workings: 1) 224-1 can be factorized as (212+1)(212-1) 2) 212-1 can be further factorized as (26+1)(26-1) 3) Therefore, 224-1 = (212+1)(26+1)(26-1) 4) The numbers between 60 and 70 that divide (26+1)(26-1) are 65 and 63. So the two numbers that exactly divide 224-1 and lie between 60 and 70 are 65 and 63.

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Real number by G R Ahmed of KVK
Euclid’s Division Lemma If there two positive integers a and b, there exist unique integer q and r satisfying a=bq+r 0≤r<b. Through this lemma the formation of the fundamental theorem of arithmetic took place and Euclid’s division algorithm is based on this lemma.
Fundamental Theorem of Arithmetic In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. It can be represented as a product of primes.
Euclid’s Division Lemma: Given positive integers a and b,there exist unique integers q and r satisfying a = bq + r, 0 ≤r < b. Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero. Euclid’s division Algorithm:- To obtain the HCF of two positive integers say a and b, with a>b, follow the steps below: Step I: Apply Euclid’s division lemma, to a and b, so we find whole numbers, and r such that a =bq+r, where 0≤ 𝑟 < 𝑏 Step II: If r=0, b is the HCF of a and b. Step III: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF
Real number by G R Ahmed of KVK
Real number by G R Ahmed of KVK
FINDING HCF BY LONG DIVISION METHOD
Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 To use Euclid's division algorithm, we apply Euclid's division lemma to given numbers c and d, to find whole numbers q and r such that a=bq+r,0≤r<b Here, a=225,b=135 225=135×1+90 Remainder is not equal to 0. Therefore, we apply the same process again on 135 and 90 135=90×1+45 Remainder is not equal to 0 again. Therefore, we apply same process again on 90 and 45. 90=45×2+0 Remainder is equal to 0. Therefore, HCF of 135 and 225 is equal to 45 which is equal
Q. 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 any positive odd integer and b=6. We can apply Euclid's division algorithm on a and b=6. ⇒a=6q+r We know that value of 0≤r<6. Therefore, all possible values of a are: a=6q =6q+1 a=6q+2 a=6q+3 a=6q+4 a=6q+5 We can ignore 6q, 6q+2 and 6q+4 because they are divisible by 2 which means they are not positive odd integers. Therefore, we are just left with 6q+1, 6q+3 and 6q+5. Therefore, any positive odd integer is of the form (6q+1) or (6q+3)or (6q+5).
Q.Any 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? Solutions: HCF of 616 and 32 would be equal to maximum number of columns in which they can march. To find HCF, we can use Euclid's division algorithm, we apply Euclid's division lemma to given numbers c and d, to find whole numbers q and r such that c=dq+r,0≤r<d Here, c=616,d=32 616=32×19+8 Remainder is not equal to 0. Therefore, we apply the same process again on 32 and 8. 32=8×4+0 Remainder is equal to 0. Therefore, HCF of 616 and 32 is equal to 8 which is equal to value of d in the last step. It means that they can march in maximum of 8 columns.
Q.Use Euclid's 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. According to Euclid's division lemma, we can say that a=3q+r, 0≤r<3 Therefore, possible values of a are: 3q, 3q+1 or 3q+2 Now, we will find cube of each one of them. (i) a³ = (3q)³=27 = 27q³= 3. 9m³=3q q=9m
(ii) (3q+1)³=27q³+1+3(3q)(3q+1) =27q³+1+27q²+9q =9(3q³ +3q²+3q)+1 let m= 9(3q³ +3q²+3q) (3q+1)³=9m + 1 (iii) (3q+2)³=(3q)³+2³+3(3q)(2)(3q+2)=27q³+8+54q²+36q =27q³+54q²+36q+8=9(3q³+6q²+4q)+8 Let m=(3q³+6q²+4q) be any integer, then we get (3q+2)³=9m+8
Real number by G R Ahmed of KVK
Q:Show that any positive odd integer is of the form 6q +1 or 6q + 3 or 6q + 5 where q is some integer We know, a = bq + r where 0 ≤ r < b By substituting b= 6 we get: a = 6q + r where[ r= 0,1,2,3,4,5] If r = 0 , a= 6q , 6q is divisible by 6 ⇒6q is even. If r = 1, a= 6q +1, 6q + 1 is not divisible by 2 If r = 2, a= 6q +2, 6q + 2 is divisible by 2 ⇒6q is even If r = 3, a= 6q +3, 6q + 3 is not divisible by 2 If r = 4, a= 6q +4, 6q + 4 is divisible by 2 ⇒6q is even If r = 5, a= 6q +5, 6q +5 is not divisible by 2 Hence the remaining integers i.e 6q + 1, 6q + 3 and 6q + 5 are odd
If n is an odd integer, then show that n2 – 1 is divisible by 8.
Q. Finds the H.C.F. of 65 and 117 and express it in the form of 65m+117n. The HCF 65 and 117 by Euclid division Lemma is 13 13 = 65 – 52 13 = 65 – (117 – 65 ) ⇒13 = 65 × 2 – 117 ⇒13 = 65 × 2 + 117 × (–1) =65m + 117n In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1
Q. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x. Ans: Using Euclid’s Division Lemma a= bq+r , 0 ≤ 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
Express the HCF of 48 and 18 as a linear combination. A=bq+r, where o ≤ r < b 48=18x2+12 18=12x1+6 12=6x2+0 ∴ HCF (18,48) = 6 Now 6= 18- (48-36) 6= 18 +36 -48 6= 54+48 6= 18x3-48x1 6= 18x3+48x(-1) i.e. 6= 18x +48y 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 Hence, x and y are not unique.
Q. Prove that any 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
Q.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 ∴ HCF (17,527) = 17 ∴ HCF (391, 425 and 527) = 17
Q. Show that the square of any positive odd integer is in the form 8k +1. Let a is an odd positive integer Therefore a=2m + 1 Squaring both sides we get a² = 4m² + 4m + 1 =4m(m + 1) + 1 We know product of two consecutive numbers is always even m(m+1)=2k a²=4(2k)+1 a² = 8 k + 1 Hence proved
Q. Prove that no number of the type 4k+2 can be a perfect square. Ans: Given the type of Number is 4k+2 = 2(2k+1) 2 is a factor of 4k+2 but 2k+1 is odd and cannot have factor 2, so 4k+2 is not divisible by 4, and therefore cannot be a perfect square.
Q.Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. 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
Q.Find all positive integral values of n for which n²+96 is perfect square. Answer: Let n² + 96 = x² ⇒ x² – n² = 96 ⇒ (x – n) (x + n) = 96 ⇒ both x and n must be odd or both even on these condition the cases are x – n = 2, x + n = 48 x – n = 4, x + n = 24 x – n = 6, x + n = 16 x – n = 8, x + n = 12 and the solution of these equations can be given as x = 25, n = 23 x = 14, n = 10 x = 11, n = 5 x = 10, n = 2 So, the required values of n are 23, 10, 5, and 2.
Real number by G R Ahmed of KVK
Real number by G R Ahmed of KVK
Q. Let a, b, c, and p be rational numbers such that p is not a perfect cube and a +b1/3 +cp2/3 =0 then show that a=b=c
Q. Check whether 6n can end with the digit 0 for any natural number n. If any digit has last digit 0 that means It is divisible by 10 And the factors of 10 = 2x5 So value 6n should be divisible by 2 and 5 both 6 is divisible by 2 but not divisible by 5 So it can not end with 0.
Example : Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Solution : If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero.
Theorem: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer. Proof : Let the prime factorization of a be as follows : a = p1.P2.P3………Pn where p1,p2…. . . ., pn are primes, not necessarily distinct. a²=( p1.p2…..pn) ( p1.p2…..pn) Now, it is given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a² Theorem: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form , p/q where p and q are co -prime, and the prime factorization of q is of the form 2ⁿ5m , where n, m are non-negative integers.
Show that 571 is a prime number.
Q.If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y. (Ans:5, -2 (Not unique)
Q.Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
What are the two numbers lying between 60 and 70, each of which exactly divides 224-1? 224-1 =( 212+1)( 212-1) =( 212+1)(26+1)(26-1) ==( 212+1) 65 Hence numbers are 65 ,63

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Real number by G R Ahmed of KVK

  • 2. Euclid’s Division Lemma If there two positive integers a and b, there exist unique integer q and r satisfying a=bq+r 0≤r<b. Through this lemma the formation of the fundamental theorem of arithmetic took place and Euclid’s division algorithm is based on this lemma.
  • 3. Fundamental Theorem of Arithmetic In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. It can be represented as a product of primes.
  • 4. Euclid’s Division Lemma: Given positive integers a and b,there exist unique integers q and r satisfying a = bq + r, 0 ≤r < b. Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero. Euclid’s division Algorithm:- To obtain the HCF of two positive integers say a and b, with a>b, follow the steps below: Step I: Apply Euclid’s division lemma, to a and b, so we find whole numbers, and r such that a =bq+r, where 0≤ 𝑟 < 𝑏 Step II: If r=0, b is the HCF of a and b. Step III: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF
  • 7. FINDING HCF BY LONG DIVISION METHOD
  • 8. Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 To use Euclid's division algorithm, we apply Euclid's division lemma to given numbers c and d, to find whole numbers q and r such that a=bq+r,0≤r<b Here, a=225,b=135 225=135×1+90 Remainder is not equal to 0. Therefore, we apply the same process again on 135 and 90 135=90×1+45 Remainder is not equal to 0 again. Therefore, we apply same process again on 90 and 45. 90=45×2+0 Remainder is equal to 0. Therefore, HCF of 135 and 225 is equal to 45 which is equal
  • 9. Q. 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 any positive odd integer and b=6. We can apply Euclid's division algorithm on a and b=6. ⇒a=6q+r We know that value of 0≤r<6. Therefore, all possible values of a are: a=6q =6q+1 a=6q+2 a=6q+3 a=6q+4 a=6q+5 We can ignore 6q, 6q+2 and 6q+4 because they are divisible by 2 which means they are not positive odd integers. Therefore, we are just left with 6q+1, 6q+3 and 6q+5. Therefore, any positive odd integer is of the form (6q+1) or (6q+3)or (6q+5).
  • 10. Q.Any 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? Solutions: HCF of 616 and 32 would be equal to maximum number of columns in which they can march. To find HCF, we can use Euclid's division algorithm, we apply Euclid's division lemma to given numbers c and d, to find whole numbers q and r such that c=dq+r,0≤r<d Here, c=616,d=32 616=32×19+8 Remainder is not equal to 0. Therefore, we apply the same process again on 32 and 8. 32=8×4+0 Remainder is equal to 0. Therefore, HCF of 616 and 32 is equal to 8 which is equal to value of d in the last step. It means that they can march in maximum of 8 columns.
  • 11. Q.Use Euclid's 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. According to Euclid's division lemma, we can say that a=3q+r, 0≤r<3 Therefore, possible values of a are: 3q, 3q+1 or 3q+2 Now, we will find cube of each one of them. (i) a³ = (3q)³=27 = 27q³= 3. 9m³=3q q=9m
  • 12. (ii) (3q+1)³=27q³+1+3(3q)(3q+1) =27q³+1+27q²+9q =9(3q³ +3q²+3q)+1 let m= 9(3q³ +3q²+3q) (3q+1)³=9m + 1 (iii) (3q+2)³=(3q)³+2³+3(3q)(2)(3q+2)=27q³+8+54q²+36q =27q³+54q²+36q+8=9(3q³+6q²+4q)+8 Let m=(3q³+6q²+4q) be any integer, then we get (3q+2)³=9m+8
  • 14. Q:Show that any positive odd integer is of the form 6q +1 or 6q + 3 or 6q + 5 where q is some integer We know, a = bq + r where 0 ≤ r < b By substituting b= 6 we get: a = 6q + r where[ r= 0,1,2,3,4,5] If r = 0 , a= 6q , 6q is divisible by 6 ⇒6q is even. If r = 1, a= 6q +1, 6q + 1 is not divisible by 2 If r = 2, a= 6q +2, 6q + 2 is divisible by 2 ⇒6q is even If r = 3, a= 6q +3, 6q + 3 is not divisible by 2 If r = 4, a= 6q +4, 6q + 4 is divisible by 2 ⇒6q is even If r = 5, a= 6q +5, 6q +5 is not divisible by 2 Hence the remaining integers i.e 6q + 1, 6q + 3 and 6q + 5 are odd
  • 15. If n is an odd integer, then show that n2 – 1 is divisible by 8.
  • 16. Q. Finds the H.C.F. of 65 and 117 and express it in the form of 65m+117n. The HCF 65 and 117 by Euclid division Lemma is 13 13 = 65 – 52 13 = 65 – (117 – 65 ) ⇒13 = 65 × 2 – 117 ⇒13 = 65 × 2 + 117 × (–1) =65m + 117n In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1
  • 17. Q. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x. Ans: Using Euclid’s Division Lemma a= bq+r , 0 ≤ 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
  • 18. Express the HCF of 48 and 18 as a linear combination. A=bq+r, where o ≤ r < b 48=18x2+12 18=12x1+6 12=6x2+0 ∴ HCF (18,48) = 6 Now 6= 18- (48-36) 6= 18 +36 -48 6= 54+48 6= 18x3-48x1 6= 18x3+48x(-1) i.e. 6= 18x +48y 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 Hence, x and y are not unique.
  • 19. Q. Prove that any 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
  • 20. Q.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 ∴ HCF (17,527) = 17 ∴ HCF (391, 425 and 527) = 17
  • 21. Q. Show that the square of any positive odd integer is in the form 8k +1. Let a is an odd positive integer Therefore a=2m + 1 Squaring both sides we get a² = 4m² + 4m + 1 =4m(m + 1) + 1 We know product of two consecutive numbers is always even m(m+1)=2k a²=4(2k)+1 a² = 8 k + 1 Hence proved
  • 22. Q. Prove that no number of the type 4k+2 can be a perfect square. Ans: Given the type of Number is 4k+2 = 2(2k+1) 2 is a factor of 4k+2 but 2k+1 is odd and cannot have factor 2, so 4k+2 is not divisible by 4, and therefore cannot be a perfect square.
  • 23. Q.Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. 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
  • 24. Q.Find all positive integral values of n for which n²+96 is perfect square. Answer: Let n² + 96 = x² ⇒ x² – n² = 96 ⇒ (x – n) (x + n) = 96 ⇒ both x and n must be odd or both even on these condition the cases are x – n = 2, x + n = 48 x – n = 4, x + n = 24 x – n = 6, x + n = 16 x – n = 8, x + n = 12 and the solution of these equations can be given as x = 25, n = 23 x = 14, n = 10 x = 11, n = 5 x = 10, n = 2 So, the required values of n are 23, 10, 5, and 2.
  • 27. Q. Let a, b, c, and p be rational numbers such that p is not a perfect cube and a +b1/3 +cp2/3 =0 then show that a=b=c
  • 28. Q. Check whether 6n can end with the digit 0 for any natural number n. If any digit has last digit 0 that means It is divisible by 10 And the factors of 10 = 2x5 So value 6n should be divisible by 2 and 5 both 6 is divisible by 2 but not divisible by 5 So it can not end with 0.
  • 29. Example : Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Solution : If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero.
  • 30. Theorem: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer. Proof : Let the prime factorization of a be as follows : a = p1.P2.P3………Pn where p1,p2…. . . ., pn are primes, not necessarily distinct. a²=( p1.p2…..pn) ( p1.p2…..pn) Now, it is given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a² Theorem: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form , p/q where p and q are co -prime, and the prime factorization of q is of the form 2ⁿ5m , where n, m are non-negative integers.
  • 31. Show that 571 is a prime number.
  • 32. Q.If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y. (Ans:5, -2 (Not unique)
  • 33. Q.Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
  • 34. What are the two numbers lying between 60 and 70, each of which exactly divides 224-1? 224-1 =( 212+1)( 212-1) =( 212+1)(26+1)(26-1) ==( 212+1) 65 Hence numbers are 65 ,63