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Q.

The units digit in the integral part of the expansion of  (15+220)19(1+(15+220)63)  is ___  

(Here integral part means the greatest integer less than or equal to that number)

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answer is 9.

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Detailed Solution

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Let  (15+220)19=I+f,0<f<1 (15220)19=f1,0<f1<1 
Adding both,  I+1=10λ
  (15+220)82=I1+f2,0<f2<1 (15220)82=f3,0<f3<1
Adding both,  I1+1=10λ1
So, the value of  I+I1  will be of the type  10λ32
But, we also get  f+f2>1

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