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Question

If (2+1)6+(21)6=198, then the integral part of (2+1)6 is
(Note : the integral part here refers to the greatest integer less than or equal to the number)

A
98
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B
99
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C
197
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D
196
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Solution

The correct option is C 197
We know if 0<a<1
Then an<1, where n is a positive integer.
Thus for (21)1 which is less than 1,
(21)6<1 (21=0.41)
Now it is given that
(2+1)6+(21)6=198
(2+1)6=198(21)6<198
Also, as shown above (21)6<1
Thus, (2+1)6=198(21)6>197
Therefore, the greatest integer for (2+1)6 is 197.

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