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Let the sum    3+33+333+……………+333333….3 {3 repeated 200 times}

be …….zyx in the decimal system i.e. x is the unit’s digit,y is the ten’s digit and so on.What is z?

a ) 0   b ) 9   c ) 7   d ) 3

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We have 3+33+333+…..+333333….3(200 times)=……….zyx.

When we add 3, 200 times we get 600, i.e., we have x=0 (the last digit of 600). Now we take 60(the first two digits of 600) from 600. Adding 3, 199 times yields 597 and 597+60=657, i.e., we have y=7. Now we take 65 from 657. Adding 3, 198 times yields 594 and 594+65=659, i.e., z=9……b) is our desired result.

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