Question

By what greatest number must 86, 64 and 31 be divided so that remainder in each case will be less than the divisor by 2​

Answer 1

Given :  86, 64 and 31 be divided so that remainder in each case will be less than the divisor by 2​

To find : Greatest divisor

Solution:

Let say  Divisor  =  b

and remainder = b  - 2

86 = bx + b - 2  => 88 = b(x + 1)

64 = by + b - 2 =>  66 = b(y + 1)

31 =  bz  + b - 2 =>  33 = b(z + 1)

=> b is HCF of  88 , 66 , 33

88 = 2 * 2 * 2 * 11

66 = 2 * 3 * 11

33 = 3 * 11

11 is the HCF of   88 , 66 , 33

86 = 11 * 7  + 9

66 = 11 * 5  + 9

31  = 11 * 2 +  9

& 9 = 11 - 2    

11 is the greatest number by which 86, 64 and 31 be divided so that remainder in each case will be less than the divisor by 2​

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