Find largest number which divides 957 and 1280 leaving remainder 5 in each case
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Given that,
when 957 and 1280 are divided by required number ,leaves remainder 5 in each case.
Then, required number exactly divides,
957-5=952 and
1280-5=1275
Prime factorization of,
952=2³×7×17
1275=5²×3×17
HCF=product of common prime factors of least power.
HCF=17
Hence 17 is the required largest number which divides 957 and 1280 and leaves remainder 5 in each case.
when 957 and 1280 are divided by required number ,leaves remainder 5 in each case.
Then, required number exactly divides,
957-5=952 and
1280-5=1275
Prime factorization of,
952=2³×7×17
1275=5²×3×17
HCF=product of common prime factors of least power.
HCF=17
Hence 17 is the required largest number which divides 957 and 1280 and leaves remainder 5 in each case.
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