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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.
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
An equal number of laddoos have been placed in 3 different boxes. The laddoos in the first box were distributed among 20 children equally, the laddoos in the second box among 24 children, and those in the third box among 12 children. Not a single laddoo was leftover. Then, what was the minimum number of laddoos in the three boxes altogether?
Find the LCM:
12, 15
Find the LCM:
105, 195
Find the HCF and LCM:
14, 28
Find the HCF and LCM:
17, 102, 170
Find the LCM:
24, 40, 80, 120
Find the LCM of the numbers given below:
42, 63
The HCF of two number is 15 and their product is 1650. Find their LCM.
The LCM of two co-primes is the sum of the numbers
Find the LCM set of numbers using prime factorisation method.
8, 12
