Find the largest number which when divided by 825 and 975 leaves remainder 5 in each case
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Answered by
19
The numbers that are exactly divisible are 825 - 5 = 800 and 975 - 5 = 970
Prime factorization of 800 = 2×2×2×2×2×5×5
Prime factorization of 970 =2×5×97
HCF(800, 970) = 2×5=10
Therefore the largest number which divides 825 and 975 leaving remainder 5 in each case is 10
Prime factorization of 800 = 2×2×2×2×2×5×5
Prime factorization of 970 =2×5×97
HCF(800, 970) = 2×5=10
Therefore the largest number which divides 825 and 975 leaving remainder 5 in each case is 10
Answered by
13
The number that are exactly divisible are 825 -5 = 800 and 975 -5 = 970
Prime factorization of 800 = 2*2*2*2*2*5*5
Prime factorization of 970 = 2*5*97
HCF (800,970) = 2*5 = 10
Therefore the largest number which divides 825 and 975 leaving remainder 5 in each case is 10.
Prime factorization of 800 = 2*2*2*2*2*5*5
Prime factorization of 970 = 2*5*97
HCF (800,970) = 2*5 = 10
Therefore the largest number which divides 825 and 975 leaving remainder 5 in each case is 10.
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