If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. One
B. Two
C. Three
D. Seven
E. Ten


Above expression could be written as d = 1/ 10(2x5)^3 x 5^4. Hence there will be 3 zeros and 3 non zeros i.e d=1/625000. IMO C.

GID09 wrote:If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. One
B. Two
C. Three
D. Seven
E. Ten


Above expression could be written as d = 1/ 10(2x5)^3 x 5^4. Hence there will be 3 zeros and 3 non zeros i.e d=1/625000. IMO C.

No not right!

I stand corrected, 1/625000 is 0.000016..so its 2.

1/(2^3 * 5^7) = 1/8 * 1/ 5^7 = .125 / 125 * 1/5^4 = .001 * .2 * .2 *.2 * .2 = .000...016

d = 1/(2^3 * 5^ 7)

If a number is divided or multiplied by a power of 10 it will only change the decimal by inserting a 0 at appropriate place..

In this case 2^3 and 5^3 will yield 1000 so questions reduces to

d = 1/(5^4)

now if we multiply denominator by 16 it yields again a number which is a power of 10.

so question reduces to d = 16

so answer is 2 i.e B

aspiregmat wrote:d = 1/(2^3 * 5^ 7)

If a number is divided or multiplied by a power of 10 it will only change the decimal by inserting a 0 at appropriate place..

In this case 2^3 and 5^3 will yield 1000 so questions reduces to

d = 1/(5^4)

now if we multiply denominator by 16 it yields again a number which is a power of 10.

so question reduces to d = 16

so answer is 2 i.e B

any more solutions??

fruti_yum wrote:If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. One
B. Two
C. Three
D. Seven
E. Ten


anyone know any approaches to tackle this question?

hey fruti,


as far as tackling this type of problem...the gmat expects you to recognize 1/n where n=1 to 9...this is something you can choose to memorize. in this case we have 1/8 which is .125. Then if you isolate 1/n where n equals 5 to various powers a pattern emerges very quickly:

1/5=.2 or 2 x 10^-1
1/25=.04 or 4 x10^-2
1/125=.008 or 8 x 10^-3
etc
you can intuit that 1/5^7 will be will have will be 128 x 10^-7....you can ignore the powers of 10. and multiply 128 by .125 and you will get a 2 digit number, 16. --->therefore b.

again the approach for problem like this is pattern recognition based on powers of a number and dividing into 1. its somewhat tedious. but its worth it if you are shooting for 700+.

you got this!

fruti_yum wrote:If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. One
B. Two
C. Three
D. Seven
E. Ten


anyone know any approaches to tackle this question?

1/(2^3*5^7)= 1/(1000)*625
=1000/(1000)(1000)(625)

1000/625=1.6
answer 2
B

1/(2³5�) =>

1/(2³5³5�) =>

1/10³ * 1/5� =>

1/1000 * 1/25 * 1/25 =>

.001 * .04 * .04

The result will have as many decimal places as everything we're multiplying together, or 3 + 2 + 2 decimal places, or 7. Since 1 * 4 * 4 = 16, and those numbers come at the end, we're left with .0000016, or five leading zeros and two nonzero digits.

fruti_yum wrote:If d = 1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. One
B. Two
C. Three
D. Seven
E. Ten

Since actually dividing 1/(2^3*5^7) would be time-consuming, we want to manipulate d so that we are working with a cleaner denominator. The easiest way to do that is to multiply d by a value that will produce a perfect power of 10 in the denominator. This means that the number of 2s in the denominator will equal the number of 5s in the denominator.

Thus, we can multiply 1/(2^3*5^7) by 2^4/2^4. This gives us:

2^4/(2^7*5^7)

2^4/10^7

16/10^7

16/10,000,000

We can stop here because we know that the 10,000,000 in the denominator means to move the decimal point after the 16 seven places to the left. The final value of d will be 0.0000016. (Note that the division of 16 by 10,000,000 did not produce any additional nonzero digits, and in general, for any positive integers a and n, a/10^n has the same number of nonzero digits as a.) Thus, d has 2 nonzero digits.

Answer: B