Faculty : Vishal Sharma More than 16 years of experience in CAT training Trained 10000+ students till date

Topic for the Session Concept of HCF/GCD Part 1(Basics)

What do you mean by HCF? Factor A number which divides a given number. Eg: 6-1,2,3,6 . .F-Highest Common Factor Eg: Highest number which divides the given number. HCF of 5 and 25 is 5 (Student's usually asks me when do we apply HCF and when LCM.This knowledge helps while solving questions). (PCS) Q)What is the HCF of '0' and '6' Ans)

HCF - Basics So one thing is clear that HCF will divide given How to find out HCF! Well two ways 1. Factorization Method 2. Division Method

HCF Prime Facorisation Method First Understand Factorisation: 18 3*6 2*9 Prime Factorisation: 18 2*9 2*3*3 breaking into lowest possible prime factors So, What is the HCF of 16,24,40 16 2*2* 2* 2 For Finding HCF we need to look at 24 2*2*2*3 1)common factors 40 = 2*2*2*5 2)Their lowest power So HCF 2A3 8

HCF Division Method When it is difficult to prime factorize number we can use Division Method. For eg take 18 & 24 1. Divide 24 by 18 -Rem 6 2. In step 2 Reminder become Divisor Divisor of last step become Dividend We repeat this process till we get Remainder 0 When we get Rem = 0, That particular divisor is the HCF

Additional Concept - HCP 1. HCF of two Co-prime numbers is 1 Eg: 5,7 & 9,10 2. HCF of two numbers of the form 'a' and 'ka' is 'a'. Eg : 4,16 HCF is 4 3. HCF of fractions a/b,c/d, e/f is equal to (HCF of Numerator term)/LCM of Denominator terms) Eg: (HCF of (a,c,e) l/LCM of (b,d,f)

Practice Questions Q)The product of two numbers is 2028 and their HCF is 13.The number of such pairs is? (PCS) Sol) Let no be 13a and 13b where a,b are co prime Now 13a* 13b 169ab 169ab- 2028 therefore ab 12 12- (1,12) (2,6) (3,4) Therefore numbers are 13,13*12 & 39,52

Q)Three sets of English,Maths,Science books containing 336,240,96 books are to be stacked in such a way that all the books are stored subjectwise and the height of each stack is same.Total number of stacks will be (SSC 2007) Sol) HCF of (336,240,96) 336 = 2^4 *3*7 240 2A4*3*5 96 2A5*3 HCF 2A4 3 48 336/48 +240/48 + 96/48 7+5+2 = 14 Number of stacks

Q)Four metal rods of length 78cm,104cm,117cm and 169cm are to be cut into parts of equal lengths.Each part must be as long as possible.What is the maximum number of pieces that can be cut off Sol) You need HCF of 78,104,117,169 78 2*3*13 104 2A3 *13 117 3A2*13 169 13A2 So HCF 13 Pieces-78/13 + 104/13 + 117/13 + 169/13 6 6 8 +9 13-36 pieces

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Vishal Sharma

I am an engineer by education and teacher by choice.I have been training students for CAT exams for 16 years.I have worked with top brands l

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Ashok Kumar

3 months ago

very nice

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