Indonesia Open Mathematical Contest

Algebra Level 3

2 4 + 2 11 + 2 15 + 2 16 + 2 21 + 2 24 4 = ? \large \sqrt[4]{2^4+2^{11}+2^{15}+2^{16}+2^{21}+2^{24}} = \ ?


The answer is 66.

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3 solutions

Chew-Seong Cheong
Jun 28, 2016

Relevant wiki: Simplifying Expressions with Radicals - Intermediate

X = 2 4 + 2 11 + 2 15 + 2 16 + 2 21 + 2 24 4 = 2 1 + 2 7 + 2 11 + 2 12 + 2 17 + 2 20 4 Assuming the form of ( 1 + x ) 4 x 4 = 2 20 x = 2 5 = 32 = 2 1 + 4 ( 2 5 ) + ( 2 + 4 ) ( 2 5 ) 2 + 4 ( 2 5 ) 3 + ( 2 5 ) 4 4 = 2 1 + 4 ( 2 5 ) + 6 ( 2 5 ) 2 + 4 ( 2 5 ) 3 + ( 2 5 ) 4 4 Note that ( 1 + x ) 4 = 1 + 4 x + 6 x 2 + 4 x 3 + x 4 = 2 ( 1 + 2 5 ) 4 4 = 2 ( 1 + 2 5 ) = 2 ( 1 + 32 ) = 66 \begin{aligned} X & = \sqrt [4]{2^4+2^{11}+2^{15}+2^{16}+2^{21}+2^{24}} \\ & = 2\sqrt [4]{\color{#3D99F6}{1+2^{7}+2^{11}+2^{12}+2^{17}+2^{20}}} \quad \quad \small \color{#3D99F6}{\text{Assuming the form of }(1+x)^4 \implies x^4 = 2^{20} \implies x = 2^5 =32} \\ & = 2\sqrt [4]{\color{#3D99F6}{1+4(2^{5})+(2+4)(2^{5})^2+4(2^{5})^3+(2^{5})^4}} \\ & = 2\sqrt [4]{\color{#3D99F6}{1+4(2^{5})+6(2^{5})^2+4(2^{5})^3+(2^{5})^4}} \quad \quad \small \color{#3D99F6}{\text{Note that }(1+x)^4=1+4x+6x^2+4x^3+x^4} \\ & = 2\sqrt [4]{(1+2^5)^4} = 2(1+2^5) = 2(1+32) = \boxed{66} \end{aligned}

Moderator note:

Good recognition of how to simplify the expression by identifying the binomial expression.

Same method The question was too beautiful

Ravi Dwivedi - 4 years, 11 months ago

Can you break down the jump from the third to the fourth step please?

Matt Bartlett - 4 years, 11 months ago

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I have added a line. Hope that it helps.

Chew-Seong Cheong - 4 years, 11 months ago

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That's great, thank you

Matt Bartlett - 4 years, 11 months ago

Lets simplify the part inside the radical first:

2 4 + 2 11 + 2 15 + 2 16 + 2 21 + 2 24 2^{4}+2^{11}+2^{15}+2^{16}+2^{21}+2^{24}

= 2 4 ( 1 + 2 7 + 2 11 + 2 12 + 2 17 + 2 20 ) =2^{4}(1+2^{7}+2^{11}+2^{12}+2^{17}+2^{20})

= 2 4 [ 1 + 2 7 + 2 11 + 2 17 + ( 2 12 + 2 20 ) ] =2^{4}[1+2^{7}+2^{11}+2^{17}+(2^{12}+2^{20})]

= 2 4 [ 1 + 2 7 + 2 11 + 2 17 + ( 2 6 + 2 10 ) 2 2 17 ] =2^{4}[1+2^{7}+2^{11}+2^{17}+(2^{6}+2^{10})^{2}-2^{17}]

= 2 4 [ 1 2 + ( 2 6 + 2 10 ) 2 + 2 7 + 2 11 ] =2^{4}[1^{2}+(2^{6}+2^{10})^{2}+2^{7}+2^{11}]

= 2 4 [ ( 1 + 2 6 + 2 10 ) 2 2 7 2 11 + 2 7 + 2 11 ] =2^{4}[(1+2^{6}+2^{10})^{2}-2^{7}-2^{11}+2^{7}+2^{11}]

= 2 4 [ 1 2 + 2 5 + ( 2 × 1 × 2 5 ) ] 2 =2^{4}[1^{2}+2^{5}+(2 \times 1 \times 2^{5})]^{2}

= 2 4 [ 1 + 2 5 ] 4 = ( 2 × 33 ) 4 = 6 6 4 =2^{4}[1+2^{5}]^{4}=(2 \times 33)^{4}=66^{4}

Now, putting it under the radical, we have 66 66 as the final answer.

For 2 4 + 2 11 + 2 15 + 2 16 + 2 21 + 2 24 4 \sqrt[4]{2^4+2^{11}+2^{15}+2^{16}+2^{21}+2^{24}}

Here's the steps:

2 4 + 2 11 + 2 15 + 2 16 + 2 21 + 2 24 4 = 16 + 2048 + 32768 + 65536 + 2097152 + 16777216 4 = 18974736 4 = 66 \begin{aligned} \sqrt[4]{ \color{#D61F06}{2^{4}} + \color{#EC7300}{2^{11}} + \color{#CEBB00}{2^{15}} + \color{#20A900}{2^{16}}+ \color{#3D99F6}{2^{21}}+\color{#302B94}{2^{24}}} &= \sqrt[4]{ \color{#D61F06}{16}+ \color{#EC7300}{2048}+\color{#CEBB00}{32768}+\color{#20A900}{65536}+\color{#3D99F6}{2097152}+\color{#302B94}{16777216} } \\&= \sqrt[4]{18974736} \\&= 66 \space \square \end{aligned}

ADIOS!!! \LARGE \text{ADIOS!!!}

Do that without a calculator in 2 minutes.

Ayush BANERJEE - 1 year, 5 months ago

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