If a=2b=6c for positive integers a, b, and c, which of the f

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[GMAT math practice question]

If a=2b=6c for positive integers a, b, and c, which of the following could be the value of abc?

A. 1000
B. 1230
C. 2250
D. 2367
E. 2488

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by [email protected] Revolution » Thu May 23, 2019 12:21 am
=>

Since 2b = 6c, we have b = 3c. Thus, abc = (6c)*(3c)*c = 18c^3 and abc is a multiple of 18. This implies that abc is a multiple of both 2 and 9.

Of the above answer choices, only 2250 is a multiple of both 2 and 9.

Therefore, the answer is C.
Answer: C

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by deloitte247 » Wed May 29, 2019 1:04 am
a = 2b = 6c
definitely,
a = 6c
2b = 6c
Hence
$$b=\frac{6c}{2}=3c$$
The value of abc = a*b*c = (6c)*(3c)*(c)
$$=18c^3$$
Therefore,
The product of abc must if be divisible by 18 without remainder
$$\frac{1000}{18}=55.56$$
$$\frac{1230}{18}=68.33$$
$$\frac{2250}{18}=125$$
$$\frac{2367}{18}=131.5$$
$$\frac{2488}{18}=138.22$$
2250 is the only option that can be divided by 18 without remainder.

$$answer\ \ is\ Option\ C\ $$

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by [email protected] » Sat Jun 01, 2019 5:27 am
[email protected] Revolution wrote:[GMAT math practice question]

If a=2b=6c for positive integers a, b, and c, which of the following could be the value of abc?

A. 1000
B. 1230
C. 2250
D. 2367
E. 2488
Let's express a and b in terms of c.

We see that a = 6c and since 2b = 6c, we see that b = 3c.

The product abc can thus be re-expressed as (6c)*(3c)*(c) = 18c^3. Thus, the product abc must be a multiple of 2 and 9. We can immediately rule out choice D, as it is odd. Of the remaining choices, we check to see which is a multiple of 9. To check a number to see if it is a multiple of 9, we add the digits of the number, and if the sum of the digits is divisible by 9, then the number itself is divisible by 9. Let's check each answer choice:

A. 1 + 0 + 0 + 0 = 1. The sum is not divisible by 9.

B. 1 + 2 + 3 + 0 = 6. The sum is not divisible by 9.

C. 2 + 2 + 5 + 0 = 9. The sum is divisible by 9.

E. 2 + 4 + 8 + 8 = 22. The sum is not divisible by 9.

Answer: C

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