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29 Dec 2022, 08:14
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If A is a positive Integer, which of the following must be divisible by 12?
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
Please help gmatphobia chetan2u
29 Dec 2022, 08:56
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PriyamRathor
If A is a positive Integer, which of the following must be divisible by 12?
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
To determine whether a number or term is divisible by 12, we must ensure that the number / term is divisible by 4 and is divisible by 3.
1. \(A^4 -A^2\)
\(A^2(A^2-1)\)
\(A^2(A+1)(A-1)\)
\(A*(A-1)A(A+1)\)
Check for divisibility by 4:
Case 1: A is even
As we have two even terms in the product, the number is divisible by 4.
Case 2: A is odd
When A is odd, (A-1) & (A+1) ⇒ even
As we have two even terms in the product, the number is divisible by 4.
Check for divisibility by 3: We have product of three consecutive terms, therefore the number will be divisible by 3.
Conclusion: As \(A^4 -A^2\) is divisible by both 4 and by 3, it is divisible by 12
2. \((A-6)^2 (A-4)(A+13)^2\)
IMO, checking the divisibility of the second and the third term is a bit tricky. We can use the knowledge of remainders and even and odd integers to do so.
We will use the concept of even / odd to check the divisibility of 4, and the concept of remainders to check the divisibility of 3.
Check for divisibility by 4:
Case 1 : A is even
even^2 * even * odd^2
As we have two even terms in the product, we can be sure that the number is divisible by 4.
Case 2 : A is odd
odd^2 * odd * even^2
Again, we see that the product has two even terms in the product, hence the number is divisible by 4
In both the above cases, the number is divisible by 4.
Check for divisibility by 3:
We can use the concept of remainders to check the divisibility of 3. "A" when divided by 3, can have a remainder 0, 1 or 2. Let's check if the term is divisible by 3 in each of the case.
Case 1: Remainder(A/3) = 0; i.e. A is a multiple of 3
If A is a multiple of 3, \((A-6)^2\) will be divisible by 3. Hence the entire term is divisible by 3.
Case 2: Remainder(A/3) = 1; i.e. A when divided by 3, leaves a remainder = 1
Remainder(\(\frac{(A-6)^2 (A-4)(A+13)^2 }{ 3} \)) = \((1-0)^2 (1-1)(1+1)^2\) = 0
In this case (A-4) will be divisible by 3, hence the entire term is divisible by 3.
Case 3: Remainder(A/3) = 2; i.e. A when divided by 3, leaves a remainder = 2
Remainder(\(\frac{(A-6)^2 (A-4)(A+13)^2 }{ 3} \)) = \((2-0)^2 (2-1)(2+1)^2\)
In this case, the last term \((A+13)^2\) will be divisible by 3.
In all the above cases, the number is divisible by 3.
Conclusion:As \( (A-6)^2 (A-4)(A+13)^2\) is divisible by both 4 and by 3, it is divisible by 12
3. \(2A (A-10) (A+13)\)
We can use the same approach in this as well. We can use the concept of even and odd to determine the divisibility by 4 and use the concept of remainders to determine the divisibility by 3
Check for divisibility by 4:
Case 1: A is even
If A is even, 2A is divisible by 4. Hence the entire term is divisible by 4.
Case 2: A is odd
If A is odd, (A+13) is even. We already have 2 in the equation, so the presence of an even term guarantees that the term will be divisible by 4.
Hence in both the cases, the number is divisible by 4.
Check for divisibility by 3:
As in previous case, we can use the concept of remainder to check the divisibility of 3.
Case 1: Remainder(A/3) = 0; i.e. A is a multiple of 3
2A (A-10) (A+13) is divisible by 3.
Case 2: Remainder(A/3) = 1; i.e. A when divided by 3, leaves a remainder = 1
Remainder (\(\frac{2A(A-10)(A+13)}{3}\)) = 2*1*(1-1)(1+1)
In this case the term (A-10) will be divisible by 3.
Case 3: Remainder(A/3) = 2; i.e. A when divided by 3, leaves a remainder = 2
Remainder (\(\frac{2A(A-10)(A+13)}{3}\)) = 2*2*(2-1)(2+1)
In this case the term (A + 13 ) will be divisible by 3.
In all the above cases, the number is divisible by 3.
Conclusion: As \(2A (A-10) (A+13)\) is divisible by both 4 and by 3, it is divisible by 12
Option D
29 Dec 2022, 20:10
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PriyamRathor
If A is a positive Integer, which of the following must be divisible by 12?
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
1. \(A^4 -A^2\)
2. \((A-6)^2 (A-4)(A+13)^2\)
3. 2A (A-10) (A+13)
A. 1 and 2 only
B. 2 and 3 only
C. 1 and 3 only
D. 1,2 and 3 only
E. None of the Above
Please help gmatphobia chetan2u
A quick method would be to substitute some values fir A say 1, 2 and 3, and check for divisibility by 12. Although you cannot be 100% sure as MUST means it should be true for all values and you cannot check for all values.
1. \(A^4 -A^2=A^2(A-1)(A+1)\)
Here A, A-1, A and A+1 are 3 consecutive integers, so multiple of 3.
If A is even, then A^2 is multiple of 4. If A is odd, (A-1)(A+1) is multiple of 4.
Hence the term is multiple of 12.
2. \((A-6)^2 (A-4)(A+13)^2\)
If A is odd, A+13 is even and (A+13)^2 is multiple of 4. If A is even (A-4)(A-6) is multiple of two even numbers and hence multiple of 4. Rather it will surely be multiple of 8.
To check for multiple of 3, add/subtract 3s multiple from each to get them closer. A-6 can be written as A, while A-4 as A-1 and A+13 as A+13-12 or A+1.
Again consecutive numbers, so multiple of 3.
3. 2A (A-10) (A+13)
If A is odd, A+13 is even and (A+13)*2 is multiple of 4. If A is odd, 2A is multiple of 4.
To check for multiple of 3, add/subtract 3s multiple from each to get them closer. A-10 can be written as A-1 and A+13 as A+13-12 or A+1.
Again consecutive numbers, so multiple of 3.
All three are multiples of 12.
D
