A
B
C
D
E
No not right!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.
any more solutions??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
hey fruti,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)*625fruti_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?
First let's examine the denominatorfruti_yum wrote:If d = 1/(2^3 x 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.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
