Question

Let's try something different

challenge to the genius

Multiply the below using exponents

27844 ×22312


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Answer 1

Answer:

To multiply:-

27844×22312

Solution:-

• First convert both units into their expanded form.

$\qquad\quad\sf \hookrightarrow (27844)\times (22312)$

$\qquad\quad\sf \hookrightarrow (20000+7000+800+40+4)(20000+2000+300+10+2)$

• now convert them into exponential form

$\qquad\quad\sf \hookrightarrow (2 \times 10^4+7 \times 10^3+8 \times 10^2+4 \times 10^1+4)(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)$

• Use distributive law

$\qquad\quad\sf \hookrightarrow 2 \times 10^4(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)+7 \times 10^3(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)+8 \times 10^2(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)+4 \times 10^1(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)+4(2 \times 10^4+2 \times 10^3+3 \times 10^2+1 \times 10^1+2)$

• Now do multiplication using law given below and construct a table like I have done below.

$\boxed{\sf a^m\times a^n=a^{m+n}}$

$\boxed {\begin {array}{ccccc}\sf 4\times 10^8 & \sf 4\times 10^7 &\sf 6\times 10^6 &\sf 2 \times 10^5 &\sf 4 \times 10^4 \\ &&&&&\\ \sf 10 \times 10^7 &\sf 14\times 10^6 &\sf 21 \times 10^5 &\sf 7 \times 10^4 &\sf 14 \times 10^3 \\ &&&&&\\ \sf 16\times 10^6 &\sf 16\times 10^5 &\sf 24\times 10^4 &\sf 8 \times 10^3 &\sf 16\times 10^2 \\ &&&&&\\ \sf 8\times 10^5 &\sf 8 \times 10^4 &\sf 12\times 10^3 &\sf 4\times 10^2 &\sf 8\times 10^1 \\ &&&&&\\ \sf 8\times 10^4 &\sf 8\times 10^3 &\sf 12 \times 10^2 &\sf 4\times 10^1 &\sf 8\end{array}}$

• Now find the units having same exponents and add them .

$\qquad\quad\sf \hookrightarrow 4 \times 10^8+18 \times 10^7+36 \times 10^6+47 \times 10^5+51 \times 10^4+42 \times 10^3+32 \times 10^2+12 \times 10^1+8$

• Convert all the units into their simplified form and write them in a box like below I have done.

$\boxed{\begin{array}{c|c|c|c|c|c|c|c|c}\\ &&&&&&&&\\ \rm 4 &\rm 0&\rm 0&\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 \\ &&&&&&&&\\ \rm 1 &\rm 8 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 \\&&&&&&&&\\ &\rm 3 &\rm 6 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 \\ &&&&&&&&\\ && \rm 4 &\rm 7 &\rm 0 &\rm 0 &\rm 0 &\rm 0 &\rm 0 \\ &&&&&&&& \\ &&&\rm 5 &\rm 1 &\rm 0 &\rm 0 &\rm 0 &\rm 0 \\ &&&&&&&& \\ &&&& \rm 4 &\rm 2 &\rm 0 &\rm 0 &\rm 0 \\ &&&&&&&&\\ &&&&&\rm 3 &\rm 2 &\rm 0 &\rm 0 \\ &&&&&&&& \\ &&&&&&\rm 1 &\rm 2 &\rm 0 \\ &&&&&&&& \\ &&&&&&&&\rm 8 \\&&&&&&&&\\ \boxed {\rm 6} &\boxed {\rm 2} &\boxed{\rm 1} &\boxed {\rm 2} &\boxed {\rm 5} &\boxed{\rm 5} &\boxed {\rm 3} &\boxed{\rm 2} &\boxed {\rm 8} \end {array}}$

$\therefore \sf{27844 \times 22312=621255328}$

Steps to do this:-

• Expand the two units first by which 0 should be at the last of it.

Example:-

11111=10000+1000+100+10+1

• now convert them into exponential form having coefficients as 10

Example:-

$\sf 11111=1 \times 10^4+1 \times 10^3+1 \times 10^2+1 \times 10^1+1$

• Now use distributive law .

$\boxed{\sf (a+b)(c+d)=a (c+d)+b (c+d)}$

• Now create boxes as I created above and get your answer.

Where can we use this?

• when we have to multiply higher digits without calculator errorlessly then we can use this .
• In time limited exams you can use this.
• I am sure it will help you in exam.

Answer 2

$\mapsto \large \tt27844 \times 22312 \\ \\\mapsto \tt (2² × 6961) × (2³ × 2789) \\ \\\mapsto \large \rm {2}^{ \tt(2 + 3)} \times 6961 \times 2789 \\ \\ \mapsto \large \rm {2}^{ \tt5} \times 6961 \times 2789 \\ \\\mapsto \large \tt 32 \times 6961 \times 2789$

• That's enough (◔‿◔) ಠ‿ಠ

Surrendered !!