The document explains applications of the Least Common Multiple (LCM) and Highest Common Factor (HCF) in various mathematical problems. It provides examples for finding maximum weights, divisibility conditions, and specific remainders when dividing numbers. The overall focus is on using LCM and HCF to solve problems involving common divisors and remainders.

1. Applications of LCM & HCF
• Greatest Number that will leave no remainder when divides a, b and c
Required number = HCF of a, b and c
Example
A shopkeeper has three cakes of weight 18 kg, 45 kg and 36 kg. If he
wants to make these cakes into pieces of equal weight without
wastage, what is the maximum possible weight of each piece?
(1) 1 (2) 9 (3) 12 (4) 13

2. Applications of LCM & HCF
• The greatest number that will divide a, b & c leaving
remainder x, y & z respectively
Required number = HCF of {(a – x), (b – y), (c – z)}
Example
What is the greatest number that divide 20, 50, and 40
leaving 2, 5 and 4 as remainder?
Here a–x=
b–y=
c–z=
Required Number =

3. Applications of LCM & HCF
• To find the greatest number that will divide x, y and z leaving the
same remainder “r” in each case.
Required number = HCF of (x – r), (y – r) and (z – r)
Example
Find the greatest number which will divide 369, 449, 689,
5009 and 729 so as to leave the remainder 9 in each case
(A) 2 (B) 49 (C) 35 (D) 40
Required Number is
HCF of {(369-9), (449-9), (689-9), (5009 – 9), (729 – 9)}
= HCF of {360, 440, 680, 5000, 720}
= 40

4. Applications of LCM & HCF
• To find the greatest number that will divide x, y and z
leaving the same remainder in each case.
Required number = HCF of |x – y|, |y – z| and |z – x|
Example
What is the greatest number that will divide 1305, 4665 and
6905 leaving in each case the same remainder? Also
calculate the remainder.
(1) 1210, 158 (2) 1120, 158
(3) 1120, 185 (4) 1210, 185

5. Solution
Here we have
|4665 – 1305|, |6905 – 4665|, |6905 – 1305|
= 3360, 2240, 5600
= 1120 x 3, 1120 x 2, 1120 x 5
Required number = HCF of (3360, 2240, 5600) = 1120
To calculate Remainder
1305 = 1120 x 1 + 185
So the Remainder = 185

6. Applications of LCM & HCF
• Least number which is exactly divisible by a, b, c
Required number = LCM of x, y and z
Sequence = n x LCM
(Where n is a natural Number)
Example
Find the greatest number of five digits which is divisible by
32, 36, 40, 42 and 48
(a) 999720 (b) 90702 (C) 90720 (d) 90730

7. Solution
Step I
LCM of (32, 36, 40, 42, 48) = 10080
Step II
Find greater number of 5 digit which multiple of 10080
10080 99999 9
90720
9279
Greatest number = 90720

8. Applications of LCM & HCF
• To find least number which when divided by a, b, c leaves
“r” as a remainder
Required number = LCM of (a, b, c) + r
Example
What is the smallest sum which a person can have such that
when he distributed @ Rs. 2.5 or Rs 20 or Rs 12 or Rs. 7.5 per
person in a group, he is always left with Rs. 2.00?
(A) Rs. 62 (B) Rs. 80 (C) Rs. 90 (D) Rs. 100

9. Solution
According to the problem the person must have the money
equal to the LCM of 2.5, 20, 12 and 7.5 and the remainder
money always left.

10. Applications of LCM & HCF
• Least number which when divided by a, b, c leaves x, y, z
as remainder, such that a – x = b – y = c – z = k (say)
Required number = LCM of (a, b, c) – k
Example
What is the greatest number of 4 digits that when divided
by the numbers 6, 9, 12, 17 leaves 5, 8, 11, 16 as
remainders respectively?
(a) 9791 (b) 9793 (c) 9895 (d) 9497

11. Solution
Step 1 Here,
6 – 5 = 9 – 8 = 12 – 11= 17 – 16 = 1 = K (let)
Step 2 LCM of 6, 9, 12, 17 = 17 x 9 x 4 = 17 x 36 = 612
Step 3 Find Greatest 4 digit number which is multiple of 612
612 9999
612 16
3879
Greatest Number
3672
= 9999 – 207 = 9792
207
Step 4 Required number = (LCM) n - K
= 9792 - 1 = 9791

12. Thank you !!