Question

Which of the following is a factor of $(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)({2}^{16} + 1)$?

(a) $2^4 - 4$
(b) $2^4 - 1$
(c) $2^8 + 8$
(d) $2^2 + 2$

Class – IXth [ANTHE] ​

Answer 1

Answer :-

$\\\;{\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}}$

Here the concept of Algebraic Identities has been used. We see we are given the identity. This can be solved by Binomial Theorem or Geometric Progression. But here to solve it in easier method we can do it by using Algebraic Identities. First we will split the terms in the required forms and then we will do the answer.

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★ Solution :-

Given,

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;(2\;+\;1)(2^{2}\;+\;1)(2^{3}\;+\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}$

Then,

Let us assume that the given equation is equal to N.

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2\;+\;1)(2^{2}\;+\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

Now multiplying and dividing the first term that is (2 + 1) by (2 - 1), we get

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{\dfrac{(2\;-\;1)}{(2\;-\;1)}\:(2\;+\;1)(2^{2}\;+\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

Now using identity, (a + b)(a - b) = a² - b²

Then,

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{\dfrac{(2^{2}\;-\;1^{2})}{1}\:(2^{2}\;+\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2^{2}\;-\;1)(2^{2}\;+\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

Again using the same identity, (a² - b²)(a² + b²) = a⁴ - b⁴ , in first two terms, we get,

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2^{4}\;-\;1^{4})(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2^{4}\;-\;1)(2^{4}\;+\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

Now again using the same identity with higher exponents, we get,

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2^{8}\;-\;1^{8})(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

$\\\;\;\;\sf{:\Rightarrow\;\;N\;=\;\bf{(2^{8}\;-\;1)(2^{8}\;+\;1)(2^{16}\;+\;1)}}$

Again using that same identity, we get,

$\\\;\;\;\sf{:\mapsto\;\;N\;=\;\bf{(2^{16}\;-\;1^{16})(2^{16}\;+\;1)}}$

$\\\;\;\;\sf{:\mapsto\;\;N\;=\;\bf{(2^{16}\;-\;1)(2^{16}\;+\;1)}}$

$\\\;\;\;\sf{:\mapsto\;\;N\;=\;\bf{(2^{32}\;-\;1^{32})}}$

$\\\;\;\;\sf{:\mapsto\;\;N\;=\;\bf{(2^{32}\;-\;1)}}$

$\\\;\;\underline{\boxed{\tt{\;\;N\;=\;\bf{(2^{32}\;-\;1)}}}}$

Now we can reduce the terms to get the correct answer by using same identity. We can notice that what all terms we removed, are factors only (2³² - 1) can be simplified into it.

$\\\;\;\sf{:\Longrightarrow\;\;(2^{32}\;-\;1)\;=\;\bf{(2^{4}\;-\;1)(2^{4}\;+\;1)(2^{8}\;-\;1)(2^{16}\;-\;1)}}$

Since, all these terms when multiplied along with identity, gives us the required simplied answer.

And factor theorem states that all the numbers which are multiplied to give a certain number are factors of that number.

Thus,

$\\\;\;\sf{:\mapsto\;\;Factors\;of\;(2^{32}\;-\;1)\;=\;\bf{(2^{4}\;-\;1),\:(2^{4}\;+\;1),\:(2^{8}\;-\;1),\:(2^{16}\;-\;1)}}$

From given option only, 2⁴ - 1 is the correct option.

So the correct option is Option b.) 2⁴ - 1.

$\\\;\large{\underline{\underline{\rm{Thus,\;factor\;of\;the\;given\;equation\;is\;\;\boxed{\bf{2^{4}\;-\;1}}}}}}$

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★ More Identities to Know :-

$\\\;\sf{\leadsto\;\;(a\;+\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;2ab}$

$\\\;\sf{\leadsto\;\;(a\;-\;b)^{2}\;=\;a^{2}\;+\;b^{2}\;-\;2ab}$

$\\\;\sf{\leadsto\;\;(a\;+\;b\;+\;c)^{2}\;=\;a^{2}\;+\;b^{2}\;+\;c^{2}\;+\;2ab\;+\;2bc\;+\;2ac}$

$\\\;\sf{\leadsto\;\;a^{2}\;+\;b^{2}\;=\;(a\;+\;b)^{2}\;-\;2ab}$

$\\\;\sf{\leadsto\;\;(a\;+\;b)^{3}\;=\;a^{3}\;+\;b^{3}\;+\;3ab(a\;+\;b)}$

$\\\;\sf{\leadsto\;\;a^{3}\;+\;b^{3}\;=\;(a\;+\;b)[a^{2}\;-\;ab\;+\;b^{2}]}$

$\\\;\sf{\leadsto\;\;a^{3}\;-\;b^{3}\;=\;(a\;-\;b)[a^{2}\;+\;ab\;+\;b^{2}]}$

$\\\;\sf{\leadsto\;\;(x\;+\;a)(x\;+\;b)\;=\;x^{2}\;+\;(a\;+\;b)x\;+\;ab}$