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getzgetzu
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 06 May 2006, 05:12
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How many non-zero digits does the decimal 1/(2^k*5^b) have? Where k and b are positive integers and k<b

1) k+b=10
2) 3<b-k<7



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laxieqv
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 06 May 2006, 08:12
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getzgetzu
How many non-zero digits does the decimal 1/(2^k*5^b) have? Where k and b are positive integers and k<b

1) k+b=10
2) 3<b-k<7
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1/(2^k*5^b) = 1/ (10^k ) * 1/ 5^ (b-k)= 1/10^k * 0.2 ^ (b-k) =
1/ 10^(k+1) * 2 ^ ( b-k) ---> the number of non-zero digits is determined by 2^ ( b-k)

1) b+k = 10, b> k ---> (b,k) = {( 9,1) , (8,2), (7,3) , (6,4)} --> b-k= { 8,6,4, 2} ---> 2^(b-k) can be one-non-zero digit such as 4 or 2-non-zero number such as 16 ---> the number of non-zero digits can't be determined --> insuff

2) b-k can be 4,5,6 , 2^( b-k) is always 2-non-zero number --> suff

B it is.
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Professor
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 25 May 2006, 18:38
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laxieqv
getzgetzu
How many non-zero digits does the decimal 1/(2^k*5^b) have? Where k and b are positive integers and k<b

1) k+b=10
2) 3<b-k<7


1/(2^k*5^b) = 1/ (10^k ) * 1/ 5^ (b-k)= 1/10^k * 0.2 ^ (b-k) =
1/ 10^(k+1) * 2 ^ ( b-k) ---> the number of non-zero digits is determined by 2^ ( b-k)

1) b+k = 10, b> k ---> (b,k) = {( 9,1) , (8,2), (7,3) , (6,4)} --> b-k= { 8,6,4, 2} ---> 2^(b-k) can be one-non-zero digit such as 4 or 2-non-zero number such as 16 ---> the number of non-zero digits can't be determined --> insuff

2) b-k can be 4,5,6 , 2^( b-k) is always 2-non-zero number --> suff

B it is.
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great work by laxi....agree with B.
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giddi77
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 25 May 2006, 20:45
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Looks confusing, but not at all after reading Laxie's solution!
Should be B!
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deowl
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How many non-zero digits does the decimal 1/(2^k*5^b) have? Updated on: 26 May 2006, 07:10
Originally posted by deowl on 25 May 2006, 22:55.
Last edited by deowl on 26 May 2006, 07:10, edited 1 time in total.
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We are asked for the number of non-zero digits in 1 / (10^(k+1) * 2 ^ ( b-k)), not in 2 ^ ( b-k) alone.

Obviously the number of such digits will be different when we divide by 16 or by 64.

So E it is.
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jaynayak
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 25 May 2006, 23:28
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laxieqv
getzgetzu
How many non-zero digits does the decimal 1/(2^k*5^b) have? Where k and b are positive integers and k<b

1) k+b=10
2) 3<b-k<7


1/(2^k*5^b) = 1/ (10^k ) * 1/ 5^ (b-k)= 1/10^k * 0.2 ^ (b-k) =
1/ 10^(k+1) * 2 ^ ( b-k) ---> the number of non-zero digits is determined by 2^ ( b-k)

1) b+k = 10, b> k ---> (b,k) = {( 9,1) , (8,2), (7,3) , (6,4)} --> b-k= { 8,6,4, 2} ---> 2^(b-k) can be one-non-zero digit such as 4 or 2-non-zero number such as 16 ---> the number of non-zero digits can't be determined --> insuff

2) b-k can be 4,5,6 , 2^( b-k) is always 2-non-zero number --> suff

B it is.
Show more


Agree with 1 is not suff, but don't agree with 2 is suff.

From your equation 1/ 10^(k+1) * 2 ^ ( b-k)
Now we know from 2) that b-k can be 4,5,6 Now 1/16 has 3 non zero digits while 1/64 has 5 non zero digits.

Hence E.
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mahesh004
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How many non-zero digits does the decimal 1/(2^k*5^b) have? 26 May 2006, 03:33
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Agree with Laxie.

Jaynayak - 2^ ( b-k) is not in the denominator. So question of 1/16 Vs 1/64 wont arise...



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