2 thoughts on “what is the greatest number that can divide each of 635, 851, 887 leaving a remainder of 5 in each case?”
Answer:
To find the greatest number which when divides 261, 933, 1391 leaving remainder 5 in each case. First we subtract the remainder from the given numbers and then calculate the HCF of New numbers by prime factorization method.
•HCF – HCF of two or more numbers = product of the smallest power of each common prime factor involved in the numbers.
SOLUTION:
GIVEN numbers are 261, 933, 1391 and remainder is 5 in each case. Then new numbers after Subtracting remainder are •261.
– 5 = 256, 933 – 5 = 998, 1391 – 5 = 1386 New numbers are 256, 998 & 1386 . HCF of 256, 998 & 1386 .
Answer:
To find the greatest number which when divides 261, 933, 1391 leaving remainder 5 in each case. First we subtract the remainder from the given numbers and then calculate the HCF of New numbers by prime factorization method.
•HCF – HCF of two or more numbers = product of the smallest power of each common prime factor involved in the numbers.
SOLUTION:
GIVEN numbers are 261, 933, 1391 and remainder is 5 in each case. Then new numbers after Subtracting remainder are •261.
– 5 = 256, 933 – 5 = 998, 1391 – 5 = 1386 New numbers are 256, 998 & 1386 . HCF of 256, 998 & 1386 .
HCF by prime factorization method :
= 2X2X2X2X2X2X2X2 = 2A8
998 = 2 x 499 = 2′ x 4991
1386 = 2x3x3x7x11 = 21 x 32x Ixill
HCF(256, 998,1386) = 21 = 2
Hence, the largest number is 2.
Step-by-step explanation:
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