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Question
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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Solution
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.
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What do you observe in the results obtained?
The LCM of 2 and 5 is:
