Qa04 hcf & lcm
6 likes2,349 views
This document provides models and examples for calculating the highest common factor (HCF), lowest common multiple (LCM), and properties of exponents and factorials. It includes 3 models for finding the LCM and HCF of numbers. It also provides rules for determining the last digit of powers of a number, and finding the largest power of a prime number that divides a factorial.
More Related Content
Similar to Qa04 hcf & lcm (20)
More from Veeraragavan Subramaniam (20)
Qa04 hcf & lcm
- 1. QA04 HCF & LCM
- 3. LCM – MODEL 1 • Any number which when divided by p,q or r leaving the same remainder s in each case will be of the form • k (LCM of p, q and r)+ s where k = 0,1,2… • If we take k = 0, then we get the smallest such number. • Example: • The least number which when divided by 5,6,7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: • K(LCM of 5,6,7 and 8) + 3 = 840k + 3 • Least value of k for which 840k + 3 divisible by 9 is • K=2. • Required number is 1683.
- 4. LCM MODEL 2 • Any number which when divided by p,q or r leaving respective remainders of s, t and u where (p–s) = (q – t) = ( r – u) -= v (say) will be of the form • K(LCM OF P, Q AND R) – V • The smallest such number will be obtained by substituting k = 1. • Example: • Find the smallest number which when divided by 4 and 7 gives remainders of 2 and 5 respectively. • LCM OF 4 AND 7 IS 28. • HENCE 28 – 2 = 26.
- 5. LCM MODEL 3 • Find the smallest number which when divided by 7 leaves a remainder of 6 and when divided by 11 leaves a remainder of 8. • The required number will be 11k + 8 • When divided by 7 leaves a remainder 6. • Subtracting 6 from 11k + 8 we have 11k + 2 which should be multiple of 7. • By trial, when k =3, we get 35. • Hence Required number is when k = 3, 11k+8 = 41.
- 6. HCF MODEL 1 • The largest number with which the numbers p,q or r are divided giving remainder of s, t and u respectively will be the HCF of the three numbers of the form (p – s), (q – t) and (r – u) • Example • Find the largest number with which when 906 and 650 are divided they leave remainders 3 and 5. • The HCF of (906 – 3) and ( 650 – 5). • HCF of 903 and 645 is 129.
- 7. HCF MODEL 2 • The largest number which when we divide by the numbers p,q and r , the remainders are the same is • HCF of (p – r) and (q – r) where r is the smallest among the three. • Example • Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in case. • Required number = H.C.F of (91 – 43), and (183 – 43) = H.C.F of 48 and 140 is 4.
- 8. LAST DIGIT OF ANY POWER • Last digit of 21 is 2 • Last digit of 22 is 4 • Last digit of 23 is 8 • Last digit of 24 is 6 • Last digit of 25 is 2 • Last digit of 31 is 3 • Last digit of 32 is 9 • Last digit of 33 is 7 • Last digit of 34 is 1 • Last digit of 35 is 3
- 9. Digit s Powers 1 1 2 3 4 5 6 7 8 9 2 2 4 8 6 2 4 8 6 2 3 3 9 7 1 3 9 7 1 3 4 4 6 4 6 4 6 4 6 4 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 7 7 9 3 1 7 9 3 1 7 8 8 4 2 6 8 4 2 6 8 9 9 1 9 1 9 1 9 1 9 For every digit unit place digits of increasing powers repeat after 4th power. This means unit place digit for power=5 is same as unit place digit for power=1 for every number. 2) For digits 2, 4 & 8 any power will have either 2 or 4 or 6 or 8 at unit place. 3) For digits 3 & 7 any power will have either 1 or 3 or 7 or 9 at unit place. 4) For digit 9 any power will have either 1 or 9 at unit place. 5) And for digits 5 & 6 every power will have 5 & 6 at unit place respectively.
- 10. LARGEST POWER OF A NUMBER IN N! • Find the largest power of 5 that can divide 216! without leaving any remainder. (or) • Find the largest power of 5 contained in 216! • Add all the quotients to get 43 + 8 + 1 = 52. • Therefore 552 is the highest power of 5 contained in 216! 5 216 Number given 5 43 Quotient 1 5 8 Quotient 2 1 Quotient 3 Please note that this method is applicable only when the number whose largest power is to be found out is a prime number. If it is not a prime number, then split the number as product of primes and find the largest power of each factor. Then the smallest amoung the largest poser of these relative factors of the given number will the largest power required.
- 11. an – bn • It is always divisible by a – b. • When it is even it is also divisible by a + b. • When it is odd it is not divisible by a + b.
- 12. an + bn • It is never divisible by a – b. • When it is odd it is also divisible by a + b. • When it is even it is not divisible by a + b.
- 13. • There are three departments having students 64,58,24 .In an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only (No mixing of departments. Find the minimum number rooms required ? • The HCF is 2. Hence 32 + 29 + 12= 73.