If a and b are positive integers, and a

Question

If  a  and  b  are positive integers, and  a  <  b , does the decimal expansion of  \frac{a}{b}  terminate?

1) The largest positive integer which is a divisor of both  a  and  b  is 24

2) The smallest positive integer which is divisible by both  a  and  b  is 6000

JerryAtDreamScore · Answer

Statement 1 Alone:

a and b are both multiples of 24, so  \frac{a}{b}  would simplify into some fraction that we don't know, so statement 1 is insufficient.

Statement 2 Alone:

a and b are both factors of 6000, so we could have a = 1 and b = 6000 or vice versa, leading to a nonterminating decimal or a terminating one. Thus statement 2 alone is insufficient.

Both Statements Combined:

6000 = 6 * 10^3 = 2^4*3*5^3 . Since a and b are both multiples of 24, both of them must share the only factor of 3. Thus a or b can only be different by the number of factors of 2 or 5. Then  \frac{a}{b}  will result in 2's times 5's only with various powers, which is terminating with any power. Then combined this is sufficient.

Answer: C

IanStewart · Answer

This question is from my Problem Set #2. A fraction of integers a/b will give us a terminating decimal if, when the fraction is reduced, the only prime divisors of the denominator are 2 and/or 5. So fractions like 1/4 or 1/5 or 1/100 produce terminating decimals, since they're reduced and 2 and/or 5 are the only prime divisors of their denominators, while reduced fractions like 1/3 or 1/6 or 1/700 do not produce terminating decimals, because there's a prime other than 2 or 5 in their denominators.

Statement 1 tells us the GCD and Statement 2 the LCM of the two numbers. Statement 1 is not sufficient, because the numbers a and b can be 24 and 72, say, and 24/72 = 1/3 does not terminate, or a and b can be 24 and 48, say, and 24/48 = 1/2 does terminate. Similarly Statement 2 is not sufficient because the numbers can be 2000 and 6000 or 3000 and 6000, and a/b can again equal 1/3 or 1/2, among many other possibilities.

Using both Statements, Statement 2 tells us our denominator b is some factor of 6000. So if b is divisible by any prime besides 2 or 5, it can only be divisible by one 3. That's the only reason a/b might not terminate. And from Statement 1, b is a multiple of 24, so b is certainly a multiple of 3, but Statement 1 also tells us a is divisible by 3. So in the fraction a/b, the '3' will cancel, and once the fraction is reduced the denominator can only be divisible by 2s and/or 5s, and we must have a terminating decimal.

sujoykrdatta · Answer

A fraction a/b is terminating / recurring depending on whether: 
After cancelling the common factors of a and b, b has only 2s and 5s left as factors => a/b is terminating
After cancelling the common factors of a and b, b has primes other than 2s and 5s left as factors => a/b is recurring

From 1: The GCD of a and b is 24. Thus, in a/b, the common factor 24 cancels out and the possibilities are: 
a) After cancelling 24, b has only 2s and 5s left as factors => a/b is terminating
b) After cancelling 24, b has primes other than 2s and 5s left as factors => a/b is recurring
Thus, Statement 1 is insufficient

From 2: The LCM of a and b is 6000. Thus, in a/b, you can have a possibility that: 
a) b has only 2s and 5s as factors: Example: a = 48, b = 250 => a/b is terminating: 
b) b has primes other than 2s and 5s as factors: Example: a = 16, b = 750 => a/b is recurring
Thus, Statement 2 is insufficient

Combining:

From 1, we can say that a and b are multiple of 24. Thus, let a = 24x and b = 24y
Note that x and y have no common factors since the common factor of a and b is 24

Thus, the LCM of a (= 24x) and b (= 24y) is 24xy 
=> 24xy = 6000 
=> xy = 250

Thus, x and y are factors of 250 = 5^3 x 2
Thus, x and y have no prime factors other than 2 and 5

Thus, the fraction a/b, after reducing to the lowest form (cancelling 24), will have only 2s and 5s as factors in the denominator 
This ensures that a/b MUST be terminating

Answer C

SatvikVedala · Answer

So the solution needs no '3' in the denominator

Going by statement 1 GCD(a,b) = 24 => 2^3 * 3
from this we can say, a can be 2^3 * 3 & b can be 2^3 * 3^2 (Common PF form GCD)
So a/b becomes 1/3, which is non-terminating

Going by statement 2 LCM(a,b) = 6000 => 2^4 * 3 * 5^3
from this we can say, a can be 2^4* 5^3 & b can be 2^4* 3 * 5^3 (Higher powers of each PF are considered for LCM)
So a/b becomes 1/3 again

Combining 1 & 2
GCD(a,b) = 24 => 2^3 * 3
LCM(a,b) = 6000 => 2^4 * 3 * 5^3
so we have one 3 in both a & b
Therefore a/b includes only 2's & 5's. Hence Terminating

bumpbot · Answer

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If a and b are positive integers, and a < b, does the decimal expansio

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