# HCF & LCM fundas

• HCF and LCM are among the initial ‘complicated’ stuffs we dealt with as a student. Someone taught me once that HCF is a postman who can go and meet all of the given numbers to deliver their letters. And LCM is like a postmaster to whom all the given numbers can go in case of any issue. Highest common factor (HCF), also known as GCD (Greatest common divisor) of a group of numbers is the greatest number that can exactly divide all the numbers in the group. Least Common Multiple (LCM) of a group of numbers is the smallest number which can be divided by all the numbers in the group.

Finding HCF of given numbers

Factorization Method

Find the HCF of 100, 225 and 900

Represent each number in their standard form.
100 = 22 * 52
225 = 32 * 52
900 = 22 * 32 * 52
what is common in all of them, 52.  HCF (100, 225, 900) = 25.
So 25 is the largest integer that can completely divide 100, 225 and 900.

Division Method:

To find the HCF of two given numbers, divide the largest by the smallest number, and then divide the dividend by the remainder. Repeat this until remainder is zero. The last dividend is the HCF of the two numbers.

Find HCF of 27 & 36

Step1: Divide 36/27, remainder = 9
Step2: Divide dividend by remainder
27/9, remainder = 0
So the HCF or GCD (greatest common Divisor) of 27 and 36 is 9

To find the HCF of three given numbers

Calculate the HCF of 2 numbers, then HCF of the result with 3rd number.

Find the GCD of 100, 225, & 900 by Successive Division Method

Phase1: Find GCD of two numbers (here 900 and 100)
Step 1: Divide the largest number 900 by the smallest number 100. And this division gives us a remainder 0 and the divisor is the GCD of 100 and 900 = 100.
Phase2: Find the GCD of 100(GCD of 900 & 100) & 225(given number)
Step 1: Divide the remaining given number 225 by the GCD of 100 & 900 i.e.100.
Step 2: Division in Step 1 gives us remainder 25. Divide the dividend (100) again by remainder (25), which gives us a remainder 0 and the divisor (25) is our GCD.

GCD of 100, 225 and 900 = 25

Finding LCM of given numbers

Factorization method

Find the LCM of 100, 225 and 900

100 = 22 * 52
225 = 32 *52
900 = 22 * 32 * 52

Write down all the prime factors that appear at least once in any of the number and then raise it to their highest available power. Product of them will give you the LCM. Using this logic in our above problem, LCM = 22 * 32 * 52 = 900.  Which means 900 is the smallest number which can be completely divided by the numbers 100, 225 and 900.

Division method

Write the given numbers as shown below and divide them with the prime numbers that can exactly divide one or more of the given numbers. On division, write the quotient in each case below the number. If any number is not divisible by its respective divisor, it is to be written as such in the next line. Keep on dividing the quotient until you get 1(as quotient of all). Multiply all the divisors to get LCM of given numbers.

Find LCM of 100, 225 and 900

5 | 100     225     900
5 | 20        45      180
2 |   4         9        36
3 |   2         3         6
2 |   1         3         3
3 |   1         1         1

LCM (100, 225, 900) = 3 * 2 * 3 * 2 * 5 * 5 = 900.

We can extend division method to find GCD or LCM of any given numbers. It is much easier and time saving to use the prime factorization method as once we write the numbers in their standard form we can get the answer just by visual inspection. Only thing we need is the speed and accuracy to write the standard form of the number, which comes through practice. In case of division method for LCM a closer look will show you that we are just using standard form method in another way.

LCM for comparing fractions

Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Which one is greater 10/9 or 7/6?
LCM of denominator = 18.
Make denominator same as the LCM for the both the fractions
20/18, 21/18 second one has the highest numerator hence the highest fraction is 7/6.

HCF and LCM of decimal numbers

In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

Find the HCF and LCM of 0.12 and 0.3
Add one zero to 0.3 so that we have same number of decimal places, 0.12 and 0.30, consider these numbers without decimals i.e. 12 and 30 and find HCF (6) and LCM (60). Mark as many decimals we removed (here 2). So the answer is HCF = 0.06 and LCM = 0.6

Useful concepts

HCF (n1, n2) * LCM (n1, n2) = n1 * n2   (works only for TWO numbers)
LCM of fractions = LCM of numerators/HCF of denominators.
HCF of fractions = HCF of numerators/LCM of denominators.
Two numbers are co primes if their HCF is 1.
LCM is always greater than or equal to the biggest among the given numbers.
HCF is always less than or equal to the smallest among the given numbers.

Some regular scenarios asked from HCF / LCM

Find the GREATEST NUMBER that will exactly divide x, y, z.
Required number = H.C.F. of x, y and z.

Find the GREATEST NUMBER that will divide x, y & z leaving remainders a, b and c respectively.
Required number = H.C.F. of (x – a), (y – b) and (z – c).

Find the LEAST NUMBER which is exactly divisible by x, y and z.
Required number = L.C.M. of x, y and z

Find the LEAST NUMBER which when divided by x, y and z leaves the remainders a, b and c respectively. It is always observed that (x – a) = (z – b) = (z – c) = K (say).
Required number = (L.C.M. of x, y and z) – K.

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.

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).

Try out the above formulae by taking some arbitrary numbers and also try to understand why it works. And most importantly share them too.In our exams, we may not find a problem directly asking to find the HCF and LCM. But questions where we need to apply the concept of HCF and LCM are very common.

The front wheels of a toy truck are 4 inches in circumference. The back wheels are 7 inches in circumference. If the truck travels in a straight line without slippage, how many inches will the truck have traveled when the front wheels have made 12 more rotations than the back wheels?

(A) 112   (B) 64   (C) 48   (D) 36   (E) 28

Front wheel covers 4 inches in one rotation and back wheel covers 7 inches for one rotation. Distance travelled by the truck will be an integral multiple of the LCM of 4 and 7 (distance covered by front wheel and back wheel for each of their rotation). LCM of 4 and 7 is 28 (4 and 7 are co primes hence HCF is 1. We can get LCM by multiplying the given numbers as HCF * LCM = Number1 * Number2) Only options that can be written as 28n are 112 and 28.

If 28, front wheel will make 7 rotations and back wheel will make 4 rotations.
Difference is 3. Not our answer.
If 112, front wheel will make 28 rotations and back wheel will make 16 rotations.
Difference is 12… our answer :)

If options are not available, then we can see that for each 28 inch (LCM) covered by the truck the difference between the number of rotations made by the front wheel and back wheel will be 3 more than the previous value, which is nothing but the difference of the given numbers divided by their HCF (here HCF =  1). Means for the first revolution, front wheel will make 7 rotations and backwheel will make 4 rotations. Difference is 3… for the second revolution difference will be 6 and so on… going with this trend after the truck covers 28 inches for fourth time the difference will be 4 * 3 = 12. So the required distance is 4 * 28 = 112.

Happy Learning :-)

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