Given that a, b, c, and, d are non-negative integers, is the

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for a fraction to be a terminating decimal, the denominator must contain only factors of 2 and 5. since evaluating the powers of the denominator, statement 2 says b=0 hence 3^b is eliminated and the fraction is left with only 2's and 5's as the denominator. therefore statement 2 is sufficient

Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

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Given that a, b, c, and, d are non-negative integers, is the fraction (ad)/(2^a3^b4^c5^d) a terminating decimal?

(1) d = (1 + a)(a^2 – 2a + 1)/((a – 1)(a^2 – 1))
(2) b = (1 + a)(a^2 – 2a + 1) – (a – 1)(a^2 – 1)

transforming the original condition and the question by variable approach method, in order for (ad)/(2^a3^b4^c5^d)=terminating decimal we need ad (numerator) to have 3 as a factor. Or, we would need 3^b removed from the denominator. This is because the terminating decimal can only have 2 or 5 as their prime factors in the denominator

But in case of 2), if b = 0, 3^b=3^0=1 and the denominator is (2^a)(4^c)(5^d). Therefore the prime factor of the denominator is 2 or 5. Since the condition is sufficient. the answer is B.

if any fraction it has to be terminating decimal.
denominator can have 2 or 5 or 2 and 5. it can terminate decimal.
whatever powers in 2 and 5
In question stem mentions a, b, c and d are non negative integers. It means it can be 0 and +ve integers.

In fraction ad/2^a 3 ^ b 4^c 5^d.
we can 4^c as 2 ^ 2c.
Now 2, 4,5 powers are terminating decimals. In 4 if power c value is 0 it becomes 1.
fraction to terminate ad/3^ b must be an integer. not a fraction.

in statement 1
evaluate d=1. we have to know value of b. so not sufficient.

In statement 2.
evaluate b=0.
so 3powerB 3^0 =1.
so ad/1is integer.
so option B is sufficient.




.

I'm slightly confused. If a or d = 0, is ad / (10)^n = terminating decimal? Is 0 a terminating decimal?

Can someone please solve the second statement and show the working?
b=(1+a)(a^2 -2a + 1) - (a-1)(a^2 -1 )
Thanks!

Hello

Given that b = (1 + a)(a^2 – 2a + 1) – (a – 1)(a^2 – 1)
Now, a^2 - 2a + 1 is square of (a-1) and can thus be written as (a-1)^2 or (a-1)(a-1).
a^2 - 1 can be read as a^2 - 1^2 and can thus be written as (a+1)(a-1)

So now we can rewrite b as:
b = (a+1)(a-1)(a-1) - (a-1)(a+1)(a-1)
As you can see, both terms are same and thus their difference will be 0.

Awww! I see. Thanks a lot, I'm really grateful.
What a relief!

Was confused because the theory wasn't shared.
In order for something to be a terminating decimal, it needs to be divided by ONLY prime factors 2 and 5.
Example:

2/5 -> terminates
2/2 -> terminates
2/(5*2) -> terminates
2/4 -> terminates
2/7 -> doesn't terminate

Now the problem:
We have (ad) divided by several prime factors to different exponents. The exponents largely don't matter, but it IS important to notice that we have a "3". This "3" throws off the terminating decimal requirement.

(1) this simplifies to d=1. This doesn't help because we still have a "3" in the denominator. B could be 0 which would make it terminating, but b is currently infinity. NS
(2) this simplifies to b=0. This means 3^b = 1. now we have only 2*1*4*5 in the denominator. This fits the condition for a terminating decimal. SUFF

Ans. B

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