Find the largest number which divides 245 and 1037 living remainder 5 in each case
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hey dear,
here is your answer,
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Required number = HCF of (245-5) and (1029-5) = HCF of 240 and 1024
Now, HCF of 240 and 1024 can be find out by prime factorisation method as:
240 = 2 × 2 × 2 × 2 × 3 × 5 and
1024 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2
Therefore HCF of 240 and 1024 = 2 × 2 × 2 × 2 = 16
Hence, 16 is the largest number which divides 245 and 1029 and leaves the remainder 5 in each case.
_______________
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here is your answer,
______________
Required number = HCF of (245-5) and (1029-5) = HCF of 240 and 1024
Now, HCF of 240 and 1024 can be find out by prime factorisation method as:
240 = 2 × 2 × 2 × 2 × 3 × 5 and
1024 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2
Therefore HCF of 240 and 1024 = 2 × 2 × 2 × 2 = 16
Hence, 16 is the largest number which divides 245 and 1029 and leaves the remainder 5 in each case.
_______________
# nikzz
HOPE U LIKE IT
CHEERS
☺☺
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