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If m and n are integers, is (mn + 2)(mn + 3)(mn + 4) divisible by 12?

all 3 numbers that are being multiplied are consecutive

(1) m is an even integer
let m=2x where x=1,2,3...........
x=1 (4*5*6) is divisible by 12
x=2 (6*7*8) is divisible by 12 etc
so A is sufficient

(2) m + n is an odd integer
sum is odd so one of them is even the other is odd and hence product is always even
same as above choice
so B is sufficient

so D is answer.
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Asked: If m and n are integers, is (mn + 2)(mn + 3)(mn + 4) divisible by 12?

(1) m is an even integer
m = 2k
(mn + 2)(mn + 3)(mn + 4) = (2kn+2)(2kn+3)(2kn+4) = 4(kn+1)(2kn+3)(kn+2) : Divisible by 4
Since (mn + 2), (mn + 3) & (mn + 4) are 3 consecutive integers, they are divisible by 3
(mn + 2)(mn + 3)(mn + 4) is divisible by 12.
SUFFICIENT

(2) m + n is an odd integer
One of (m,n) is even and other is odd.
Let m=2k ; n=2l+1
mn = 2k(2l+1) = 4kl + 2k
(mn + 2)(mn + 3)(mn + 4) = (4kl + 2k + 2)(4kl + 2k + 3)(4kl + 2k + 4) = 4(2kl + k + 1)(4kl + 2k + 3)(4kl + k + 2): Divisible by 4
Since (mn + 2), (mn + 3) & (mn + 4) are 3 consecutive integers, they are divisible by 3
(mn + 2)(mn + 3)(mn + 4) is divisible by 12.
SUFFICIENT

IMO D
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D .
Notice that those 3 are consecutive nos.

For a no. to be divisible by 12
it should be divisible by 3 and 4.
Since the no. is product of 3 consecutive nos. it is divible by 3 and 4 as well.

A is sufficient as we know that mn is even .( even×odd×even)

B is also sufficient.
even×odd×even.

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Every multiple of three consecutive numbers is divisible by 24, and any multiple of numbers divisible by 24 is justifiably divisible by 12
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Product of n consecutive integers is always divisible by n!

(mn + 2)(mn + 3)(mn + 4) is the product of 3 consecutive integers , hence it is always divisible by 3!
now, if 2 of the three numbers are even, it means one is a multiple of 4 ( out of 2 consecutive even integers, one is always a multiple of 4.) so the product is multiple of 2*4*3 = 24

St 1) m is even hence mn+2 and mn+4 are 2 consecutive even integers. hence so, the product is divisible by 24.
St2) m+n is odd , hence one of m and n is even , other is odd.hence mn+2 and mn+4 are 2 consecutive even integers. hence so, the product is divisible by 24.

Answer D

Bunuel
If m and n are integers, is (mn + 2)(mn + 3)(mn + 4) divisible by 12?

(1) m is an even integer
(2) m + n is an odd integer



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Hi, Bunuel , I wanted to ask that since this question already mentions in the main stem that m and n are integers wouldn't the answer to the question be yes even without the given conditions? mn will be either even or odd and so, the product of the three consecutive numbers (mn+2),(mn+3) and (mn+4) will be divisible by 4 anyway. Irrespective of the given statements, the answer will always be yes. So, in this question it seems to me that the 2 statements were unnecessary. Am I right or have I misunderstood the question? Would appreciate the help. TIA.
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sam12rawat
Hi, Bunuel , I wanted to ask that since this question already mentions in the main stem that m and n are integers wouldn't the answer to the question be yes even without the given conditions? mn will be either even or odd and so, the product of the three consecutive numbers (mn+2),(mn+3) and (mn+4) will be divisible by 4 anyway. Irrespective of the given statements, the answer will always be yes. So, in this question it seems to me that the 2 statements were unnecessary. Am I right or have I misunderstood the question? Would appreciate the help. TIA.


There are 2 possible cases:

mn is odd:
(mn+2)*(mn+3)*(mn+4) = odd*even*odd
This is definitely divisible by 2, but not necessarily by 4. It's also divisible by 3. Hence, the product is definitely divisible by 6

mn is even:
(mn+2)*(mn+3)*(mn+4) = even*odd*even
This is definitely divisible by 4. It's also divisible by 3. Hence, the product is definitely divisible by 12

This, considering both cases, we can only say th product is divisible by 6, not necessarily by 12.

So, we need the two statements.
Hope it helps!

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Thank you sujoykrdatta for your response. I now understand where I went wrong :)
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sam12rawat
Thank you sujoykrdatta for your response. I now understand where I went wrong :)

Also, below might help:

• Sum of \(n\) consecutive integers equals the mean multiplied by the number of terms, \(n\). Given consecutive integers \(\{-3, -2, -1, 0, 1,2\}\), \(mean=\frac{-3+2}{2}=-\frac{1}{2}\), (mean equals to the average of the first and last terms), so the sum equals to \(-\frac{1}{2}*6=-3\).

• If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(n\) consecutive integers is always divisible by \(n!\). Given \(n=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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