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rsrighosh
I am getting answer as B

The reason is:-
as \(a,b,c,d\) are all positive integers and consecutive number in ascending order

then \(\frac{abc}{d}\) =\(\frac{ (a)(a+1)(a+2)}{(a+3)}\) is obviously never an integer.
How C is valid

Please help Bunuel

Have you tried plugging some numbers? For example, a = 3?
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Kritisood
Is abc/d an integer if a, b, c, and d are positive integers?

(1) (a + b + c)/d is an integer.

(2) {a, b, c, d} are consecutive integers and arranged in ascending order

target find whether abc/d an integer
given a,b,c,d are +ve integers
#1
(a + b + c)/d is an integer.
let a,b,c = 1 and d= 3 ; yes
and a,b,c = 1 and d = 4 ; no
insufficient
#2
{a, b, c, d} are consecutive integers and arranged in ascending order
1,2,3,4 ; 1*2*3 /4 ; NO
3,4,5,6 ; 3*4*5 / 6; yes
insufficient
from 1 &2
3,4,5,6 ; ; 3+4+5/6 ; yes and ; 3*4*5/6 ; 10 ;
sufficient OPTION C
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Kritisood
Is abc/d an integer if a, b, c, and d are positive integers?

(1) (a + b + c)/d is an integer.

(2) {a, b, c, d} are consecutive integers and arranged in ascending order

Since each statement seems likely to be insufficient on its own, let's test when the statements are combined.

Case 1, when the statements are combined:
Since a, b, c and d are consecutive integers, a+b+c = 3b and d=b+2
Since \(\frac{a+b+c}{d} =\) integer, we get:
\(\frac{3b}{b+2} =\) integer

Thus, 3b must be a multiple of b+2, yielding the following options:
3b= b+2
3b=2b+4
3b=3b+6
3b=4b+8
And so on.

Only the option in green yields positive values for a, b, c and d:
3b=2b+4
b=4, with the result that a=3, b=4, c=5 and d=6
In this case, \(\frac{abc}{d} = \frac{60}{6} = 10\), so the answer to the question stem is YES.

Statement 1:
In Case 1, the answer to the question stem is YES.

Case 2: a=2, b=3, c=4, and d=9, with the the result that \(\frac{a + b + c}{d} = \frac{2+3+4}{9} = 1\)
In this case, \(\frac{abc}{d} = \frac{24}{9} = \frac{8}{3}\), so the answer to the question stem is NO.

Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2:
In Case 1, the answer to the question stem is YES.

Case 3: a=1, b=2, c=3, and d=4
In this case, \(\frac{abc}{d} = \frac{6}{4} = \frac{3}{2}\), so the answer to the question stem is NO.

Since the answer is YES in Case 1 but NO in Case 3, INSUFFICIENT.

Statements combined:
Only Case 1 satisfies both statements.
In Case 1, the answer to the question stem is YES.
SUFFICIENT.

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Bunuel
rsrighosh
I am getting answer as B

The reason is:-
as \(a,b,c,d\) are all positive integers and consecutive number in ascending order

then \(\frac{abc}{d}\) =\(\frac{ (a)(a+1)(a+2)}{(a+3)}\) is obviously never an integer.
How C is valid

Please help Bunuel

Have you tried plugging some numbers? For example, a = 3?

Bunuel
Hi Bunuel,

Can you please tell me where i am going wrong by taking a=-1,b=0,c=1 and d=2?
In the above case, both statements are taken and we get the answer "No".
So won't the answer to the question is E?
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Lucky1994
Question says a,b,c,d are +ve integers :)



Lucky1994
Bunuel
rsrighosh
I am getting answer as B

The reason is:-
as \(a,b,c,d\) are all positive integers and consecutive number in ascending order

then \(\frac{abc}{d}\) =\(\frac{ (a)(a+1)(a+2)}{(a+3)}\) is obviously never an integer.
How C is valid

Please help Bunuel

Have you tried plugging some numbers? For example, a = 3?

Bunuel
Hi Bunuel,

Can you please tell me where i am going wrong by taking a=-1,b=0,c=1 and d=2?
In the above case, both statements are taken and we get the answer "No".
So won't the answer to the question is E?

Posted from my mobile device
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Bunuel
rsrighosh
I am getting answer as B

The reason is:-
as \(a,b,c,d\) are all positive integers and consecutive number in ascending order

then \(\frac{abc}{d}\) =\(\frac{ (a)(a+1)(a+2)}{(a+3)}\) is obviously never an integer.
How C is valid

Please help Bunuel

Have you tried plugging some numbers? For example, a = 3?

Bunuel
Hi Bunuel,

Can you please tell me where i am going wrong by taking a=-1,b=0,c=1 and d=2?
In the above case, both statements are taken and we get the answer "No".
So won't the answer to the question is E?

Notice that we are given that a, b, c, and d are positive integers, so none of them can be -1 or 0. But even if we were not told that the variables are positive, still a = -1, b = 0, c = 1 and d = 2, would give that abc/d = 0, which IS an integer, so we'd still have an YES answer.
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Kritisood
Is abc/d an integer if a, b, c, and d are positive integers?

(1) (a + b + c)/d is an integer.

(2) {a, b, c, d} are consecutive integers and arranged in ascending order

What can you say..
1) a, b, c and d are positive integers
2) If d is prime and the greater than the other three, answer will surely be NO.
3) If d is a factor of any of the other three, answer will be YES.
4) If factors of d are also factors of product of abc, surely YES.

(1) (a + b + c)/d is an integer.
\(\frac{a+b+c}{d}=y\), where y is an integer.
But d could be a prime greater than the other 3.....a=b=c=1, and d=3......\(\frac{abc}{d}=\frac{1}{3}\)...NO
a=b=c=2, and d=2......\(\frac{abc}{d}=\frac{8}{2}=4\)...YES
Insuff

(2) {a, b, c, d} are consecutive integers and arranged in ascending order
So numbers are b-1, b, b+1, b+2...
If numbers are 1, 2, 3, 4...NO
If numbers are 3, 4, 5, 6...YES
Insuff

Combined..
Statement I tells us that \(\frac{(a + b + c)}{d}\)
=> \(\frac{(b-1)+b+(b+1)}{b+2}=x......3b=(b+2)x.......(3-x)b=2x........b=\frac{2x}{3-x}\)
Since b is positive, x can be
a) \(x=1...b=1\) so the set = 0, 1, 2, 3, but a cannot be 0
b) \(x=2..b=\frac{2*2}{(3-2)}=4\), so the set = 3, 4, 5, 6......\(\frac{abc}{d}=\frac{3*4*5}{6}=10\)..YES, an integer
No other possibility
Suff

C
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