LCM and HCF Practice Test 2

Question : 6 of 40

Marks: +1, -0

Let N be the greatest natural number that will divide 13,511, 13,903 and 14,589 leaving the same remainder in each case. The sum of digits of N is:

Solution:

We can solve this by division method
x = 13,511 y = 1,390, z = 14,589

(y-x)=13,903-13,511=392

(z-y)=14,589-13,903=686

(z-x)=14,589-13,511=1,078

HCF of 392, 686, 1,078 is 98
Sum of digits of N is 17.

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