Question

Find the greatest number that will divide 261,933,1391 and leaves remainder 5 in each case

Answer 1

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 :
256 = 2×2×2×2×2×2×2×2 = 2^8
998 = 2 × 499 = 2¹ × 499¹
1386 = 2×3×3×7×11 = 2¹ × 3²× 7¹×11¹

HCF(256, 998,1386) = 2¹ = 2

Hence, the largest number is 2.

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Answer 2

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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