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प्रश्न
Find the smallest number which when divided by 8, 9, 10, 15, or 20 gives a remainder of 5 every time.
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उत्तर
LCM of 8, 9, 10, 15, and 20 are given by
| 2 | 8, 9, 10, 15, 20 |
| 2 | 4, 9, 5, 15, 10 |
| 5 | 2, 9, 5, 15, 5 |
| 3 | 2, 9, 1, 3, 1 |
| 2, 3, 1, 1, 1 |
LCM = 2 × 2 × 5 × 3 × 2 × 3
= 20 × 6 × 3
= 120 × 3
= 360
The smallest number = 360 + 5 = 365
Hence, 365 is the smallest number which when divided by 8, 9, 10, 15, and 20 gives a remainder of 5 every time.
संबंधित प्रश्न
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
46, 51
Find the HCF and LCM:
32, 16
Find the LCM:
18, 42, 48
Find the LCM:
24, 40, 80, 120
Which of the following numbers is divisible by 3?
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
Find the LCM set of numbers using prime factorisation method.
14, 42
Find first three common multiples of the given number.
24, 16
Find L.C.M of the given number.
8 and 14
Find the LCM of the following numbers:
- 9 and 4
- 12 and 5
- 6 and 5
- 15 and 4
Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?
