6. Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case. 7. Find the largest number which divides 245 and 1029 leaving reminder 5 in each case. 8. Write the smallest digit in the blank space that will make 705-4 divisible by 3.
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Answer:
1st
Required number = (L.C.M of 15, 20, 48 and 36) + 9
= (2×2×3×5×4×3) = 720 + 9
= 729
2nd
It is given that the required number when divides 245 and 1029, the remainder is 5 in each case, This means that 245−5=240 and 1029−5=1024 are completely divisible by the required number.
It follows from this that the required number is a common factor of 240 and 1024. It is also given that the required number is the largest number satisfying the given property. Therefore, it is the HCF of 240 and 1024.
Let us now find the HCF of 240 and 1024 by Prime factorization method(refer above image).
Clearly, HCF of 240 and 1024 is common divisor i.e., 2×2×2×2=16.
Hence, required number =16.
hope it's correct and helpful for you
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