HCF and LCM

Highest Common Factor

 The largest positive integer which divides two or more integers without any reminder is called Highest Common Factor (HCF) or Greatest Common Divisor or Greatest Common Factor (GCF). Such as 12 and 18 have 1, 2, 3, and 6 as common factors or divisors but among them 6 is the highest common factor. So H.C.F. of 12 and 18 is 6.

Least Common Multiple

 The least common multiple (LCM) of two or more numbers is the smallest number (not zero) that is a multiple of these numbers. Such as 12 and 18 have 36, 72, 108, 144, …. as common multiples but 36 is the least among them. So L.C.M. of 12 and 18 is 36.

Few tips

1)      Product of two numbers = L.C.M. × H.C.F.

2)      Product of n numbers. = L.C.M of n numbers × Product of the HCF of each possible pair

3)      H.C.F. of decimal numbers

         Step 1: Find the H.C.F. of the given numbers without decimal.

         Step 2: Put the decimal point (in the H.C.F. of Step 1) from right to left according to the MAXIMUM decimal places among the                                  given numbers.

4)       L.C.M. of decimal numbers

           Step 1: Find the L.C.M. of the given numbers without decimal.

           Step 2: Put the decimal point (in the L.C.M. of Step 1) from right to left at the place equal to the MINIMUM decimal places among   the given numbers.

5)      H.C.F. of fractions =

HCF

6)       L.C.M. of fractions =

LCM

Important Results

1)      Find the GREATEST NUMBER that will exactly divide x, y, z.

Required number = H.C.F. of x, y, and z.

2)      Find the GREATEST NUMBER that will divide x, y and z leaving remainders a, b and c respectively.

Required number = H.C.F. of (x – a), (y – b) and (z – c).

3)      Find the GREATEST NUMBER that will divide x, y and z leaving the same remainder in each case.

Required number = H.C.F of (x – y), (y – z) and (z – x).

4)      Find the LEAST NUMBER which is exactly divisible by x, y and z.

Required number = L.C.M. of x, y and z (least divided).

5)      Find the LEAST NUMBER which when divided by x, y and z leaves the remainders a, b and c respectively.

Then, it is always observed that (x – a) = (z – b) = (z – c) = K (say).

∴ Required number = (L.C.M. of x, y and z) – K.

6)      Find the LEAST NUMBER which when divided by x, y and z leaves the same remainder ‘r’ each case.

Required number = (L.C.M. of x, y and z) + r.

Few Problems

1)      What is the greatest number which exactly divides 110, 154 and 242?

 Sol. The required number is the HCF of 110, 154 & 242.

110 = 2 × 5 × 11

154 = 2 × 7 × 11

242 = 2 × 11 × 11

∴ HCF = 2 × 11 = 22

 2)      What is the highest 3 digit number, which is exactly divisible by 3, 5, 6, and 7?

 Sol. The least no. which is exactly divisible by 3, 5, 6, & 7 is LCM (3, 5, 6, 7,) = 210. So, all the multiples of 210 will be exactly divisible by 3, 5, 6 and 7. Hence, greatest 3 digit number is 840. (210 × 4).

3)      Find the H.C.F. of 1.08, 0.36 and 0.9.

 Sol. Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

H.C.F. of given numbers = 0.18.

 4)      Find the least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18.

 Sol. L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4   = 364.

5)      The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Find their L.C.M.

 Sol. Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

 6)      There are some students in the class. Mr. A brought 140 sweets and distributed to the students equally, then he was left with some sweets. Mr. B brought 180 sweets and distributed equally to the students. He was also left with the same no of sweets as Mr. A was left. Mr. C brought 260 sweets, did the same thing and left with the same no of sweets. What is the max possible no. of students that were in the class?

 Sol. The question can be stated as, what is the highest number, which divides 140, 180 and 260 gives the same remainder, i.e. HCF ((180 −140),(260 −180),(260 −140)).

i.e. HCF (40, 80, 120) = 40.

You can also refer to our study materials on Syllogism and Time and Work. In case of any doubts, feel free to post your queries on our Facebook page.

Regards,

Way2bank

https://www.facebook.com/way2bank

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