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Powers and Roots Practice Test 2

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Question : 8 of 40

Marks: +1, -0

The value of

2^32 - (2 + 1) (2^2 + 1)(2^4 + 1)

(2^8 + 1 )(2^16 + 1)

is equal to:

0
1
2
$2^16$

Solution:

Concept:

Use repeated difference of squares:

(a-b)(a+b)=a^2-b^2

.

Explanation:

Let

P=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)

.

Since

2-1=1

, multiplying by

(2-1)

does not change

P

.

So

P=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)

.

Now apply the identity repeatedly:

(2-1)(2+1)=2^2-1

.

(2^2-1)(2^2+1)=2^4-1

.

(2^4-1)(2^4+1)=2^8-1

.

(2^8-1)(2^8+1)=2^{16}-1

.

(2^{16}-1)(2^{16}+1)=2^{32}-1

.

Thus

P=2^{32}-1

.

Required value

=2^{32}-(2^{32}-1)=1

.

Answer:

1

, so the correct option is B.

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