A) 4
B) 5
C) 6
D) 8
Correct Answer: A
Solution :
The greatest N = HCF of \[(1350-x),\] \[(4665-x)\] and \[(6905-x),\] where x is the remainder \[=HCF\,\,\text{of}\,\,(4665-1305),\] \[(6905-4665)\] and \[(6905-1305)\] = HCF of 3360, 2240 and 5600 Again, \[\therefore \] \[N=1120\] Sum of digits \[=1+1+2+0=4\]You need to login to perform this action.
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