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terminating decimal

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terminating decimal

by aleph777 » Sat Jan 22, 2011 4:45 pm
I'd love a little clarification behind the theory on this one:

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

If a, b, c, d, and e are non-negative integers and p=2^a3^b and q=2^c3^d5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d


OA: B
[spoiler]
I read a solution in MGMAT that says a terminating decimal can be guaranteed if the denominator can be expressed in terms of 2^x5^y, but I'd like to know a bit more about the theory that determines this.[/spoiler]
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by Night reader » Sat Jan 22, 2011 5:07 pm
aleph777 wrote:I'd love a little clarification behind the theory on this one:

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

If a, b, c, d, and e are non-negative integers and p=2^a3^b and q=2^c3^d5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d


OA: B
[spoiler]
I read a solution in MGMAT that says a terminating decimal can be guaranteed if the denominator can be expressed in terms of 2^x5^y, but I'd like to know a bit more about the theory that determines this.[/spoiler]
traditional test of terminating number would be p/q where p and q are the products of its prime numbers
2^(a-c) * 3^(b-d) / 5^e
by testing statement (1) we see that 3^(b-d) is left out, hence it can be p/3^n as we know division by 3 is problematic, i.e. not all numbers are finitely divisable by 3
st(2) b-d>0 hence we get rid of 3 in the denominator; while 2 in the denominator is not a barrier for finite division... we don't need entire MGMAT representation here of 2^n * 5^n
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by anshumishra » Sat Jan 22, 2011 5:10 pm
aleph777 wrote:I'd love a little clarification behind the theory on this one:

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.

If a, b, c, d, and e are non-negative integers and p=2^a3^b and q=2^c3^d5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d


OA: B
[spoiler]
I read a solution in MGMAT that says a terminating decimal can be guaranteed if the denominator can be expressed in terms of 2^x5^y, but I'd like to know a bit more about the theory that determines this.[/spoiler]
Theory
Lets check the theory a bit.
0.x = x/10 = x/(2*5)
0.xy = xy/100 = xy/(10*10) = xy/2^2*5^2

So whenever we try to reduce a decimal into a fraction the denominator is some power of 10 (and hence of 2*5).
Please note however that after it is further reduced you may lose some power of 2 and 5.

For example :

0.15 = 15/100 = 15/(2^2*5^2)
But after further reduction : 0.15 = 15/100 = 3/20 = 3/(2^2*5^1).

So, any number with terminating decimal will have the denominator in the form of 2^m * 5^n (where m,n >=0).


Now the problem :

p/q = 3^(b-d)/2^(c-a) * 5^(e)

Statement 1:
a> c

If b>= d then the fraction will be non-terminating , else terminating --- Not sufficient

Statement 2:
b>d
Now, the denominator is of the form 2^0*5^e, so it is terminating -- Sufficient

Hence, B
Thanks
Anshu

(Every mistake is a lesson learned )
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