fraction with terminating decimal

[vishal.pathak]

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

[shankar.ashwin]

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)

[neelgandham]

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 !

Answer : E

[LalaB]

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.

[fueledGMAT]

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!

[chaitanyabhansali]

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.