Published by:
CGP EDU Academic Team
Published on: August 12, 2026
Let R = (8 + 3 
) 20 and [R] = the greatest integer less than or equal to R. Then -
A
[R] is even
B
[R] is odd
C
R – [R] = 1 – 
D
R + R[R] = 1 – R 2
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(b, c, d)
R = [R] + f = (8 + 3 
) 20
= 20 C 0 8 20 + 20 C 1 8 19 (3 
) + …. 0 < f < 1
F = (8 – 3 
) 20
= 20 C 0 8 20 – 20 C 1 8 19 (3 
) + … 0 < F < 1
[R] + f + F = 2[ 20 C 0 8 20 + 20 C 2 8 18 (3 
) 2 + …]
= an even integer
f + F = an integer = 1 as 0 < f < 1 0 < F < 1
∴ [R] = an even integer – 1 = an odd integer
R – [R] = f = 1 – F = 1 – (8 – 3 
) 20
= 1 – 
RF = 1 = R(1 – f) = 1 = R{1 – R + [R]} = 1,
R + R[R] = 1 + R 2
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