fraction with terminating decimal

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by shankar.ashwin » Mon Nov 21, 2011 8:51 am
You get a terminating decimal only when the denominator contains factors of 2 or 5 or 2*5 (in any powers). All other numbers in the denominator produce recurring decimal.eg(3,5,7,9,11 etc)


128 is 2^7. So thats a terminating decimal.

All other options have other numbers in written in their base form (prime numbers raised to power something)

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by neelgandham » Mon Nov 21, 2011 9:06 am
If a decimal fraction terminates, then it can be written in the form a/100, a/1000, a/10000 and so on

1/(n) is a recurring decimal if n is prime and not equal to 2 or 5

A. 10/189 = 1/3 *(A fraction) = Recurring decimal * A fraction = Recurring decimal
B. 15/196 - 1/7 *(A fraction) = Recurring decimal * A fraction = Recurring decimal
C. 16/225 - 1/3 *(A fraction) = Recurring decimal * A fraction = Recurring decimal
D. 25/144 - 1/3 *(A fraction) = Recurring decimal * A fraction = Recurring decimal
E. 39/128 - Leftover :D !

Answer : E
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by LalaB » Mon Nov 21, 2011 10:19 am
vishal.pathak wrote:Which of the following fractions has a decimal equivalent that is a terminating decimal?
A. 10/189 B. 15/196 C. 16/225 D. 25/144 E. 39/128

Is there some way to solve this other than actual evaluation of every option
OA is E
my 1st thought was to find a fraction that does not have 3 in the denominator. i just summed all digits of 128 (1+2+8=11) and understood that it is not divisible by 3. as far as I know, any number that has an even number in its denominator is a terminating number.

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by fueledGMAT » Mon Nov 21, 2011 11:11 am
shankar.ashwin wrote:You get a terminating decimal only when the denominator contains factors of 2 or 5 or 2*5 (in any powers). All other numbers in the denominator produce recurring decimal.eg(3,5,7,9,11 etc)


128 is 2^7. So thats a terminating decimal.

All other options have other numbers in written in their base form (prime numbers raised to power something)
I will expand on this a bit because of how it is worded...

what shankar means to say is that it will terminate if and ONLY if the prime factorization is a combination of 2s and 5s ONLY. If you have ANY other prime factor in the answer other than a 2 or a 5, the fraction does not terminate...

i.e -
1/8 = 1/(2*2*2) = terminates
1/10 = 1/(2*5) = terminates
1/20 = 1/(2*2*5) = terminates
1/25 = 1/(5*5) = terminates
1/21 = 1/(3*7) = DOES NOT TERMINATE!!
1/60 = 1/(2*3*2*5) = DOES NOT TERMINATE
30/60 = (3*2*5)/(2*3*2*5) = 3's cancel each other out, terminates!

or for a really tough one:
82.5/100 = 825/1000 = (11*3*5*5)/(5*5*5*2*2*2) = terminates! remember, ONLY the denominator matters!

even harder one:
39/130 = (3*13)/(2*5*13) = 13s cancel each other out, only 2s and 5s in denominator... terminates!
"Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away." ~ Antoine de Saint-Exupery

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by chaitanyabhansali » Tue Nov 22, 2011 11:13 am
to be precise, when you simplify a fraction in its simplest form, then the denominator should contain only 2's and 5's as its factors.