A Huge Summation it is. #2

Number Theory Level 4

$\color{#3D99F6}{\displaystyle \sum_{n=1}^9 \left( {11 \times n} \right)^{11 \times n}}$

Find the last two digits of the above summation.

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The answer is 87.

1 solution

Mas Mus
May 30, 2015

Let's write all items of the summation:

${11}^{11}=(10+1)^{11}=100\times{C_1}+111\times{1}^{11}\\{22}^{22}=(20+2)^{22}=100\times{C_2}+221\times{2}^{22}\\{33}^{33}=(30+3)^{33}=100\times{C_3}+331\times{3}^{33}\\\vdots\\(90+9)^{99}=100\times{C_9}+991\times{9}^{99}$

It's clear that $100\times{C_1}, 100\times{C_2}, 100\times{C_3}, \ldots, 100\times{C_9}$ are divisible by $100$ , so we just need to find the reminder of $111\times{1}^{11}, 221\times{2}^{22}, 331\times{3}^{33}, \ldots, 991\times{9}^{99}\pmod{100}$ .

By using Euler's theorem, CRT, and the properties of modular arithmetic we will find that

${2}^{22}\equiv4\pmod{100}~~~~~~~{3}^{33}\equiv23\pmod{100}~~~~~{4}^{44}\equiv56\pmod{100}\\{5}^{55}\equiv25\pmod{100}~~~~~{6}^{66}\equiv56\pmod{100}~~~~~{7}^{77}\equiv7\pmod{100}\\{8}^{88}\equiv16\pmod{100}~~~~~{9}^{99}\equiv89\pmod{100}$

Now, we have

$11\times{1}+21\times{4}+31\times{23}+41\times{56}+51\times{25}+61\times{56}+71\times{7}+\\81\times{16}+91\times{89}\equiv11+84+13+96+75+16+97+96+99\\\equiv587\equiv\boxed{87}\pmod{100}$

Like that how can we find the last 3 digits?????????????

Abhisek Mohanty - 6 years ago

Please read this

Mas Mus - 6 years ago

I already read the solution for the problem you asked to see.Can you please explain it again.

Rama Devi - 6 years ago

@Rama Devi – Please reply soon

Rama Devi - 6 years ago

@Rama Devi – which part should I explain?

Mas Mus - 6 years ago

@Mas Mus – Please explain the last part

Rama Devi - 6 years ago

@Rama Devi – please explain soon.

Rama Devi - 6 years ago

@Rama Devi – ${11\choose{2}}\times{100}\times{1^{11}}+111\times{1^{11}}=55\times{100}+111=5611\equiv611\pmod{1000}$

Mas Mus - 6 years ago

@Mas Mus – But 611 is not the answer

Rama Devi - 6 years ago

@Rama Devi – I just show you how to get 611. You can find out other items by following the steps above. Now, look how to get the answer.

$611\times{1}+321\times{304}+131\times{523}+41\times{56}+51\times{125}+161\times4{56}+\\371\times{207}+681\times{616}+91\times{889}\equiv611+97584+68513+\ldots+80899\\\equiv611+584+513+296+375+416+797+496+899\\\equiv4987\equiv\boxed{987}\pmod{1000}$

Mas Mus - 6 years ago

@Mas Mus – Mas Mus, Can I ask? How could $22^{22} = 100 × C_2 + 221×2^{22}$ ? And so on?

Victor Zhang - 5 years, 7 months ago

@Victor Zhang – Using Binomial Expansion , we can write the expansion of $\left(22\right)^{22}$ as

$\begin{array}{c}&\left(20+2\right)^{22}&={22\choose0}~{20}^{22}\times{{2}^{0}}+\binom{22}{1}~{20}^{21}\times{{2}^{1}}+\binom{22}{2}~{20}^{20}\times{{2}^{2}}+\dots+\binom{22}{20}~{20}^{2}\times{{2}^{20}}+\binom{22}{21}~{20}^{1}\times{{2}^{21}}+\binom{22}{22}~{20}^{0}\times{{2}^{22}}\\\\&=100\left[{22\choose0}~{10}^{20}\times{{2}^{22}}\times{{2}^{0}}+{22\choose1}~{10}^{19}\times{{2}^{21}}\times{{2}^{1}}+{22\choose2}~{10}^{18}\times{{2}^{20}}\times{{2}^{2}}+\dots+{22\choose20}~{2}^{2}\times{{2}^{20}}\right]+220\times{{2}^{22}}+{2}^{22}\\\\&=100\times{C_2}+221\times{{2}^{22}}\end{array}$

Mas Mus - 5 years, 7 months ago

A Huge Summation it is. #2

Join the Brilliant Classes and enjoy the excellence. Also checkout Foundation Assignment #2 for JEE.

The answer is 87.

1 solution

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