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

_________________

Simple way to always control time during the quant part.

How to solve main idea questions without full understanding of RC.

660 (Q48, V33) - unpleasant surprise

740 (Q50, V40, IR3) - anti-debrief