## HCF LCM Concept, Questions & Tricks

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HCF & LCM is a very important topic for IBPS, SBI, SSC & all other Competitive examinations.

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HCF & LCM HCF (Highest Common Factor): The greatest number which can completely divide the numbers is called their highest common factor (HCF). It is also known as GCD (greatest common divisor). e.g. HCF of (24 & 32) 24 = = 32 = = 25 So HCF of 24 & 32 = 2x2x2 = 8 (just take lowest power) HCF is also determined by a division method: 24) 32 (1 24 8) 24 (3 24 x Application of HCF: Type 1: The greatest number which can divide x, y & z completely- Solution: Find the HCF of x, y & z. Examplel. Find the greatest number which can divide 286 & 616 616) = 22 completely. Solution: HCF (286 286) 616 (2 572 44 ) 286 264 22) (6 44 (2 44
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Type 2: The greatest number which when divide x, y & z leave remainders p, q & r respectively - Solution: Find HCF of {x — p, y — q, z — r}. Example 2. Find the greatest number that will divide 148, 246 and 623 leaving the remainders 4, 6 and 11 respectively. Solution: HCF (148 -4 , 246 -6 , 623 - 11 ) 144= 240 612 = so I-ICF (144, 240, 612) = 12 I-ICF (144, 240 , 612) Type 3: The greatest number which when divide x, y & z leave same remainders in each case - Solution: Find HCF of {x — y, y — z, z — x} Example 3. What is greatest number which when divide 333, 555 & 777 leaves same remainder? Solution: The greatest number which when divide 333, 555 & 777 leave same remainders - I-ICF (555 - 333, 777 - 555, 777 - 333) - HCF (222, 222, 444) = 222 LCM (Lowest Common Multiple): The least number which is completely divisible by the numbers is called their lowest common multiple (LCM) e.g. LCM of (24 & 36) = 72 24 = = 36 = so LCM of 24 & 36 = - — 72 (just take the highest power)
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Application of LCM: Type 1: The least number which is completely divisible by x, y & z - Solution : Find their LCM. Example 4. Find the least number which is completely divisible by 24, 36 & 72. Solution: 24 = - 36 = 22>02 72 = so LCM of 24, 36 & 72 = = 72 Type 2: The least number which when divide x, y & z leave same remainders k in each case - Solution: Find LCM of {x, y , z} + k. Example 5. Find the least number which when divided by 3, 4 and 5 leaves same remainder 2 in each case. Solution: The least number which when divided by 3, 4 and 5 leaves same remainder 2 in each case LCM (3, 4, 5) +2 = 60+2 = 62 Type 3: The least number which when divide x, y & z leave remainders p, q & r respectively - Solution: Find the value of x — p = y —q = z— r = k Find LCM of {x, y & z} — k Example6. Find the least number which when divided by 3, 4 and 5 leaves a remainder of 1, 2 and 3 respectively. Solution : Here k = 3-1 So the least number which when divided by 3, 4 and 5 leaves a remainder of 1, 2 and 3 respectively = LCM (3, 4 & 5) - K - 58
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Type 4: Some person start running around a circular stadium and complete one round in x seconds, Y seconds and z seconds & so on. In how much time will they meet again at the starting point? Or Few bells toll at intervals of x sec, y sec , z sec & so on. Find after what interval they will again toll together. Solution: Take the LCM of (x, y, z & ... ) Example7. A, B and C start running around a circular stadium and complete one round in 27 s, 9 s and 36 s, respectively. In how much time will they meet again at the starting point? Solution: LCM of 27, 9 and 36 = 108 So they will meet again at the starting point after 108 s. i.e., 1 min 48 s. Example8. 5 bells commence tolling together and toll at intervals 2, 4, 6, 8 and 10 seconds respectively. Find in 40 minutes, how many times do they toll together? Solution : LCM of 2, 4 , 6, 8 and 10 sec = 120 sec = 2min Hence, the bells toll together after 120 seconds = 2 min We are asked to find, how many times they will toll together in 40 min. In 40 min, bells toll together for 21 min {1 is added because in starting, 5 bells commence tolling together}

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