OG question

diegocuenca wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point? Any theories?

The digits to the right of the decimal point indicate division by a power of 10:
.1 = 1/10 = 1/10¹.
.01 = 1/100 = 1/10².
.001 = 1/1000 = 1/10³.

To determine the number of zeros that will appear to the right of the decimal point when 1/(2�5³) is put into decimal notation, factor out as many 10's as possible from the denominator:
2�5³
=2� * (2³5³)
= 64 * (2*5)³
= 64 * 10³
= 64000.

Thus, 1/(2�5³) = 1/64000 = .00001...

There are 4 zeros to the right of the decimal point.

Responding to a PM... but really just tacking on one extra bit to the above, lest anyone get mired in long division. Once we've got the 1/64000 established, we do NOT need to actually calculate the decimal value of that fraction. Rather, we can just think about the fact that 1/64,000 is somewhere between 1/100,000 and 1/10,000.

1/100,000 is .00001, so 1/64,000 must be greater than this--which lets us conclude it has at most 4 leading zeros.
1/10,000 is .0001, so 1/64,000 must be less than this--which lets us conclude it has at least 4 leading zeros.

So 4 leading zeros it is.

GMATGuruNY wrote:

diegocuenca wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point? Any theories?

The digits to the right of the decimal point indicate division by a power of 10:
.1 = 1/10 = 1/10¹.
.01 = 1/100 = 1/10².
.001 = 1/1000 = 1/10³.

To determine the number of zeros that will appear to the right of the decimal point when 1/(2�5³) is put into decimal notation, factor out as many 10's as possible from the denominator:
2�5³
=2� * (2³5³)
= 64 * (2*5)³
= 64 * 10³
= 64000.

Thus, 1/(2�5³) = 1/64000 = .00001...

There are 4 zeros to the right of the decimal point.

I read the problem as "One half to the power of 9 multiplied by five cubed." Did I make a fundamental mistake somewhere, or is this an issue where the questions on the site are not as easy to read as the questions from official materials?

edvhou812 wrote:

diegocuenca wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point? Any theories?

I read the problem as "One half to the power of 9 multiplied by five cubed." Did I make a fundamental mistake somewhere, or is this an issue where the questions on the site are not as easy to read as the questions from official materials?

Ha, well actually, as it's written, it should technically be taken as "one divided by (2 to the 9th) -- all that times (5 to the third)." But the solutions above all assume that everything after the / is in the denominator of one single fraction (with a numberator of 1). So I think it is an issue of what you said. Technically, for onto-screen transfer (assuming our interpretation is correct, we'd need parentheses around (2^9 * 5^3). (And technically, had it intended to be "one half to the power of 9," it would've needed parentheses around the (1/2).) Order of operations tends to make us insert parentheses quite often when we translate once-vertical problems into strictly-horizontal problems!

good question- here is how I figured

1/(2^9*5^3)= 1/(2^6*10^3) = 10^(-3)/64

10^(-3) gives us three zeros after decimal .000X -1st

now 1/x where 1<x<9 gives - no zeros after decimal
11<x<99 gives - 1 zero after decimal **64 qualifies in this range -2nd

so - 3 zeros from first + 1 zero from second = 4 zeros after decimal and before a non-zero digit
-Mayank

correction - 10<x<101 gives - 1 zero after decimal **64 qualifies in this range -2nd

also to note 1/x where 100<x<1001 gives - 2zeros after decimal
and so the series continues...

Awesome guys, I look everyone's point of view. Sorry about the (x) issue.

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