The integral part of (6√6+14)201 is even
The integral part of (6√6+14)201 is even
The correct option is A True
We can write (6√6+14)201 = I + f, where I is the integral part and f is the fractional
part. We want to find if I even. In most of the questions like this, we change the
sign and consider the number.
i.e.(6√6−14). This number will be positive and less than 1 in most of the cases.
(6√6−14) falls between 0 and 1.
Or 0 < (6√6−14) < 1. If we consider the positive powers also, it will lie
between 0 and 1.
i.e. 0 < (6√6−14)201 < 1. We did this because we can add/subtract this quantity with
(6√6−14)201 . There is a chance of some terms in the binomial expansion getting
cancelled and some other terms getting added. We decide if we will perform
addition or subtraction based on the question. In this case the power is odd, so
we will perform subtraction. This is because we want the terms with radical sign
to get cancelled.
Let (6√6−14)201 = f'.
(6√6+14)201 = (6√6−14) + 201C1(6√6)200 141 + 201C2(6√6)199 142 ................(1)
(6√6−14)2×100+1 = (6√6)201 + 201C1(6√6)200 141 + 201C2(6√6)199 142 ................(2)
(1) - (2) gives
(6√6+14)201 - (6√6−14)201 = 2[201C1(6√6)200(14)1 - 201C3(6√6)198(14)3............]
We have (6√6+14)201 = I + f and (6√6−14)201 = f'
⇒ I + f - f' = 2[201C1(6√6)200141 + 201C3(6√6)198143............] .............(3)
2[201C1(6√6)200141 + 201C3(6√6)198(14)3...........]is an even integer.
Now let us consider f-f' separately,
0 < f < 1 and 0 < f' < 1
f - f' will attain its maximum value when f = 1 and f' = 0 and it will attain its
minimum value when f = 0 and f' = 1.
⇒ -1 < f - f' < 1
I + f - f' is an even integer (from (3)).
I is also an integer. For I + f - f' to be an integer, f - f' should also be an integer.
The only integer value it can take is 0, because -1 < f - f' < 1 .
⇒f - f' = 0
⇒f = f'
⇒I + f - f' = I = an even integer
True
We can write (6√6+14)201 = I + f, where I is the integral part and f is the fractional
part. We want to find if I even. In most of the questions like this, we change the
sign and consider the number.
i.e.(6√6−14). This number will be positive and less than 1 in most of the cases.
(6√6−14) falls between 0 and 1.
Or 0 < (6√6−14) < 1. If we consider the positive powers also, it will lie
between 0 and 1.
i.e. 0 < (6√6−14)201 < 1. We did this because we can add/subtract this quantity with
(6√6−14)201 . There is a chance of some terms in the binomial expansion getting
cancelled and some other terms getting added. We decide if we will perform
addition or subtraction based on the question. In this case the power is odd, so
we will perform subtraction. This is because we want the terms with radical sign
to get cancelled.
Let (6√6−14)201 = f'.
(6√6+14)201 = (6√6−14) + 201C1(6√6)200 141 + 201C2(6√6)199 142 ................(1)
(6√6−14)2×100+1 = (6√6)201 + 201C1(6√6)200 141 + 201C2(6√6)199 142 ................(2)
(1) - (2) gives
(6√6+14)201 - (6√6−14)201 = 2[201C1(6√6)200(14)1 - 201C3(6√6)198(14)3............]
We have (6√6+14)201 = I + f and (6√6−14)201 = f'
⇒ I + f - f' = 2[201C1(6√6)200141 + 201C3(6√6)198143............] .............(3)
2[201C1(6√6)200141 + 201C3(6√6)198(14)3...........]is an even integer.
Now let us consider f-f' separately,
0 < f < 1 and 0 < f' < 1
f - f' will attain its maximum value when f = 1 and f' = 0 and it will attain its
minimum value when f = 0 and f' = 1.
⇒ -1 < f - f' < 1
I + f - f' is an even integer (from (3)).
I is also an integer. For I + f - f' to be an integer, f - f' should also be an integer.
The only integer value it can take is 0, because -1 < f - f' < 1 .
⇒f - f' = 0
⇒f = f'
⇒I + f - f' = I = an even integer
