mba4me Posted September 12, 2004 Share Posted September 12, 2004 If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have? Could anyone tell me how to approach this problem. Quote Link to comment Share on other sites More sharing options...
GMAT04TestTaker Posted September 12, 2004 Share Posted September 12, 2004 Multiply by 2^4 in Numerator and Denominator. So, you are left with 2^4 =>16=>2. Quote Link to comment Share on other sites More sharing options...
mba4me Posted September 12, 2004 Author Share Posted September 12, 2004 I dont understand. Can you elaborate Quote Link to comment Share on other sites More sharing options...
mba4me Posted September 12, 2004 Author Share Posted September 12, 2004 Sorry. I got it. Quote Link to comment Share on other sites More sharing options...
papakehteyhain Posted September 12, 2004 Share Posted September 12, 2004 but i don't understand it.. plz explain.. whats 2^4 ? showing 2 raise to power 4 ? and how to solve it ? Quote Link to comment Share on other sites More sharing options...
gmat168 Posted September 12, 2004 Share Posted September 12, 2004 Multiply by 2^4 in Numerator and Denominator. So, you are left with 2^4 =>16=>2. Can you elaborate more? So you're left with 2^4/10^7 Quote Link to comment Share on other sites More sharing options...
GMAT04TestTaker Posted September 12, 2004 Share Posted September 12, 2004 So you're left with 2^4/10^7 2^4/10^7 : In this fraction, denominator will only add zero digits in the final terminating decimal, which we are not supposed to count. That leaves us with numerator 2^4 = 16. Therefore number of non-zero digits are just 2. (1 and 6) Alternate => 2^4/10^7 = 16/10000000 = 0.0000016 => Non zero digits : 2 Quote Link to comment Share on other sites More sharing options...
riz1983 Posted September 12, 2004 Share Posted September 12, 2004 good explanation thanx Quote Link to comment Share on other sites More sharing options...
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