Exercise 3.5
1. Find the LCM of the following numbers by prime factorisation method.
(i) 12 and 15 (ii) 15 and 25 (iii) 14 and 21
(iv) 18 and 27 (v) 48, 56 and 72 (vi) 26, 14 and 91.
(i) 12 and 15
LCM of 12 and 15 by Prime Factorization
Prime factorization of 12 and 15 is
Hence, the LCM of 12 and 15 by prime factorization is
(ii) 15 and 25
Prime factorization of 15 and 25 is
Hence, the LCM of 15 and 25 by prime factorization is
(iii) 14 and 21
Prime factorization of 14 and 21 is
Hence, the LCM of 14 and 21 by prime factorization is
(iv) 18 and 27
Prime factorization of 18 and 27 is
Hence, the LCM of 18 and 27 by prime factorization is
(v) 48, 56 and 72
Prime factorization of 48, 56, and 72 is
Hence, the LCM of 48, 56, and 72 by prime factorization is
(vi) 26, 14 and 91
Prime factorization of 26, 14, and 91 is (2 × 13) =
Hence, the LCM of 26, 14, and 91 by prime factorization is
2. Find the LCM of the following numbers by division method.
(i) 84, 112, 196 (ii) 102, 119, 153 (iii) 45, 99, 132, 165
(i) 84, 112, 196
| Prime Factor | 84 | 112 | 196 |
|---|---|---|---|
| 2 | 42 | 56 | 98 |
| 2 | 21 | 28 | 49 |
| 3 | 7 | 28 | 49 |
| 7 | 1 | 4 | 7 |
| 2 | 1 | 2 | 7 |
| 2 | 1 | 1 | 7 |
| 7 | 1 | 1 | 1 |
LCM Calculation:
Therefore LCM(84, 112, 196) =
(ii) 102, 119, 153
| Prime Factor | 102 | 119 | 153 |
|---|---|---|---|
| 2 | 51 | 119 | 153 |
| 3 | 17 | 119 | 51 |
| 3 | 17 | 119 | 17 |
| 17 | 1 | 119 | 1 |
| 7 | 1 | 17 | 1 |
| 17 | 1 | 1 | 1 |
LCM Calculation:
Therefore, LCM(102, 119, 153) =
(iii) 45, 99, 132, 165
| Prime Factor | 45 | 99 | 132 | 165 |
|---|---|---|---|---|
| 2 | 45 | 99 | 66 | 165 |
| 2 | 45 | 99 | 33 | 165 |
| 3 | 15 | 33 | 11 | 55 |
| 3 | 5 | 11 | 11 | 55 |
| 5 | 1 | 11 | 11 | 11 |
| 11 | 1 | 1 | 11 | 11 |
| 11 | 1 | 1 | 1 | 1 |
LCM Calculation:
Therefore, LCM(45, 99, 132, 165) =
3. Find the smallest number which when added to 5 is exactly divisible by 12, 14 and 18.
12, 14, 18
| Prime Factor | 12 | 14 | 18 |
|---|---|---|---|
| 2 | 6 | 7 | 9 |
| 2 | 3 | 7 | 9 |
| 3 | 1 | 7 | 3 |
| 3 | 1 | 7 | 1 |
| 7 | 1 | 1 | 1 |
LCM Calculation
Therefore, LCM(12, 14, 18) =
LCM of 12, 14 and 18 is
Therefore, the smallest number which when added to 5 is exactly divisible by 12, 14 and 18 is
4. Find the greatest 3 digit number which when divided by 75, 45 and 60 leaves: (i) no remainder (ii) the remainder 4 in each case.
(i) 75, 45, 60
| Prime Factor | 75 | 45 | 60 |
|---|---|---|---|
| 2 | 75 | 45 | 30 |
| 2 | 75 | 45 | 15 |
| 3 | 25 | 15 | 5 |
| 3 | 25 | 5 | 5 |
| 5 | 5 | 5 | 1 |
| 5 | 1 | 1 | 1 |
LCM Calculation:
Therefore, LCM(75, 45, 60) =
ii) To get the remainder 4 when we divide the greatest 3 – digit number by 75, 45 and 60.
We have to add 4 to the greatest 3 – digit number, which is exactly divisible by 75, 45 and 60.
By adding 4 to 900 we get:
Therefore, the greatest 3 – digit number divisible by 75, 45 and 60 by leaving remainder 4 is
5. There are three measuring tapes of 64 cm, 72 cm and 96 cm. What is the least length that can be measured by any of these tapes exactly?
(i) 64 cm, 72 cm, 96 cm
| Prime Factor | 64 | 72 | 96 |
|---|---|---|---|
| 2 | 32 | 36 | 48 |
| 2 | 16 | 18 | 24 |
| 2 | 8 | 9 | 12 |
| 2 | 4 | 9 | 6 |
| 2 | 2 | 9 | 3 |
| 2 | 1 | 9 | 3 |
| 3 | 1 | 3 | 1 |
| 3 | 1 | 1 | 1 |
LCM Calculation:
Therefore, 576 cm is the least length that can be measured by any of the tape exactly.
6. Prasad and Raju met in the market on 1st of this month. Prasad goes to the market every third day and Raju goes every 4th day.On what day of the month will they meet again?
The day on which Prasad and Raju met in the market is 1st of this month.
Prasad goes to the market every 3rd day.
Raju goes to the market every 4th day.
To find the day on which they meet again, we have to find the LCM of 3 and 4.
LCM of 3 and 4 =
So, Raju and Prasad meet again after