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d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin

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d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin  [#permalink]

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New post Updated on: 18 Jul 2015, 17:07
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If d = \(\frac{1}{(2^k*5^m)}\), m > k (m and k are positive integers) is a terminating decimal, then there are how many non-zero digits in its decimal notation?

1) m + k =10
2) 3 < m – k < 7

Originally posted by Mascarfi on 18 Jul 2015, 16:35.
Last edited by ENGRTOMBA2018 on 18 Jul 2015, 17:07, edited 1 time in total.
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Re: d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin  [#permalink]

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New post 18 Jul 2015, 17:04
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Mascarfi wrote:
d = \(\frac{1}{(2^k*5^m)}\), m > k (m and k are positive integers) is a terminating decimal. Are there how many non-zero digits in its decimal notation?

1) m + k =10
2) 3 < m – k < 7


Dividing 1 by 5 in different exponents gives us increasing number of decimals and increasing exponent of 2:
1/5 = 0.2
1/25 = 0.04
1/125 = 0.008

Dividing 1 by 2 in different exponents gives us increasing number of decimals and increasing exponent of 5:
1/2 = 0.5
1/25 = 0.25
1/125 = 0.125

Dividing 1 by product of both numbers shift result to 1
1/2*5 = 0.1
1/2*2*5*5 = 0.01
1/2*2*2*5*5*5 = 0.001

So if we have more 2 in denominator then number in decimal part will be equal to exponent of 5 (exponent will be equal difference between exponent of 2 and exponent of 5)
and if we have more 5 in denominator then number in decimal part will be equal to exponent of 2 (exponent will be equal difference between exponent of 2 and exponent of 5)

So we need to know on how much differentiate exponents of m and k

1) m+k = 10 difference can be 0 --> m and k = 5 than we will have 1 in decimal part
or difference can be 6 --> m = 8 and k = 2 than we will have \(2^6 = 64\) in decimal part

2) 3 < m – k < 7
m - k can be 4, 5, 6
so we will have values from \(2^4=16\) to \(2^6=64\)
All this values have 2 nonzero digits
So sufficient

Answer is B
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d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin  [#permalink]

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New post 01 May 2017, 19:00
1. m+k=10
m could be 9, k could be 1
m 8, k 2
m 7, k3, etc etc
I'll pause to note that I've taken a page out of Bunuel's book for handling problems such as these (terminating decimals). When we have the fraction in the form 1/2^k*5^m, where k and m are distinct integers, it's easiest to multiply the top and bottom of the fraction by 2 or 5 raised to a power to get k and m equal. So if m is 9, we multiply top and bottom by 2^8. From there, we know that 2^8 = 256, and 2^8*5^8 = 10^8. So, there are three non-zero digits in the terminating decimal representation. As m decreases and k increases, there will be less non-zero digits.
1 is Insufficient.

2. Applying the same logic as above, and the fact that m and k must be either 8 and 2 or 7 and 3, respectively, we know that the numerator will become either 2^6 or 2^4. Both of these are less than 100 but greater than 10. As such, there will be 2 non-zero digits in the decimal representation of d.
2 is Sufficient - Answer B.

Edit: See the link below for Bunuel's approach that I used.
https://gmatclub.com/forum/if-d-1-2-3-5 ... l#p1052704
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Re: d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin  [#permalink]

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New post 09 Oct 2017, 11:29
Oh the language play :(
Took me a while to get it:the statement says" then there are how many non-zero digits in its decimal notation?" --> means we need to find out EXACT number of zeros; and not whether it can be determined (my big error here)

St 1: there are many case of m+ k = 10, 6&4, 7&3, 8&2, 9&1, 0&10 --> each time the number of zeros in deno. differ and hence the number of zeroes in decimals differ. So, INSUFF

St 2: 3< m -k< 7 --> k is 4,5,6 --> 8-2 or 7-3 (5 is 8 -3 or 9 -4; so not possible) so we know how many zeroes exactly would happen. (exact difference b/w m and k is known). so Suff
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Re: d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin  [#permalink]

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New post 09 Oct 2017, 11:35
Harley1980 - I had a query with this part.

2) 3 < m – k < 7
m - k can be 4, 5, 6
so we will have values from 24=1624=16 to 26=6426=64
All this values have 2 nonzero digits

So sufficient

Answer is B

Why is not 2 to power 5 counted? And why is it 2? Why not 5? Could you detail out your reasoning?
Thank you!
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Re: d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin &nbs [#permalink] 09 Oct 2017, 11:35
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