12 Days of Christmas GMAT Competition - Day 2: If n is an integer

Ans: A. I only

Out of 7, 21 and 42, we can rule out 42 immediately since 4n-1 will always be odd for all values of n and 42 is an even number

Assuming values of n=5 and 6, we can see that 4n-1 = 19 and 23 respectively. Hence this cannot be the answer as well

Now, only 7 remains to be verified

For n=2, 4n-1=7 and 8n+40=56 and for both 7 and 56, 7 is a factor

Therefore, the answer is A. I only

4n-1 8n+40 = 8(n+5) Common Factors
N + - + -
0 -1 -1 40 40 1
1 3 -5 48 32 3
2 7 -7 56 24 7
3 11 -13 64 16 1
4 15 -17 72 8 1
5 19 -21 80 0 -

The answer is C) I and II only.
This is because for n=2, 4n-1 = 7 and 8n+40 is 56. Both are divisible by 7.
And for n=16, 4n-1=63 and 8n+40=8(n+5)=8*21=168, both divisible by 21.

But, 4n-1 will always be an odd integer, and thus will not be divisible by 42.

We are tasked with determining which of the given numbers could be a factor of both 4n - 1 and 8n + 40. To solve, we evaluate the greatest common divisor of the two expressions.

The expressions are 4n - 1 and 8n + 40. Subtract twice 4n - 1 from 8n + 40 to eliminate n: (8n + 40) - 2(4n - 1) equals 8n + 40 - 8n + 2, which simplifies to 42.

The greatest common divisor of 4n - 1 and 8n + 40 must divide 42, because their linear combination equals 42.

For option I, 7 is a factor of 42, so 7 could be a factor. For option II, 21 is a factor of 42, so 21 could be a factor. For option III, 42 is the greatest common divisor, so 42 could be a factor.

The answer is all three: I, II, and III.

Let us discuss few points before look into the values given. Firstly, this is "COULD BE" question, which means the values given are not necessarily need to satisfy all possible values of n. Secondly, our answer could be a factor of both 4n-1 and 8n+40.

4n-1 : This is one less than a multiple of 4. So it is always an odd number. E.g: 15, 19, 23, 27...
8n+40: This is always a multiple of 8, because we can factor 8 out from 8n+40.

Let's discuss the choices now. Our number needs to be a factor of both 4n-1 and 8n+40.

Even number cannot be the factor for odd number. So 42 cannot be the answer. Hence D and E are not correct answers.

Now let's plug in some number for n and check whether it is possible factor.

if n = 16, 4n-1 is 63 and 8n+40 is 168. 21 is factor for both the numbers. And if 21 is a factor, then 7 is also the factor since 21 is a multiple of 7. So the answer is C.

In this case, good idea is to check with 21 first, try to think a number which is multiple of 21.

Hi everyone

Straightforward (Testing Cases)

n=2
4n-1 = 7 is divisible by 7.
8(n+5) = 8*7 is divisible by 7.
I - Correct

n=16
4n-1 = 63 is divisible by 21.
8(n+5) = 8*21 is divisible by 21.
II - Correct

For divisibility of 42 we can't find any number for n.


So the answer is C (I+II) could be a factor of the terms.

Let's suppose, 4n-1 is a multiple of 7=> 4n-1=7k, therefore, n= (7k+1)/4. now, put the value of n in 8n+40, we get-
14k+42, which is div by 7. therefore, when (4n-1) is a multiple of 7, (8n+40) is also the multiple of 7.

Similarly, 4n-1 =21k, therefore value of n=(21k+1)/4. putting this value in 8n+4, we get, 42k+42, which is also div. by 21. therefore, when 4n-1 is div by 21 so as 8n+40.

4n-1= 42k, therefore, n=(42k+1)/4, again put the value in 8n+40, we get, 84k+42, which is again div. by 42.

So, for any integral value of n, for which 4n-1 is div. by either 7,21 or 42, (8n+40) will also be div. by 7,21 and 42 as well. Therefore, opt. (E) is correct.

The expressions are 4n−14n - 14n−1 and 8n+408n + 408n+40. Their difference is:
(8n+40)−2(4n−1)=42.(8n + 40) - 2(4n - 1) = 42.(8n+40)−2(4n−1)=42.
Any common factor of 4n−14n - 14n−1 and 8n+408n + 408n+40 must divide 42.
Factors of 42: 1,2,3,6,7,14,21,421, 2, 3, 6, 7, 14, 21, 421,2,3,6,7,14,21,42.
Thus, 7,21,7, 21,7,21, and 424242 are all possible factors.
Answer: E. I, II, and III

Let A = 4n - 1, B = 8n + 40
We can express B according to A:
B = 8n + 40 = 2.(4n-1) + 42 = 2A + 42
Common factors of both must accommodate 42
--> Possible divisors of 42: 1,2,3,6,7,21,42 --> ANSWER E

If something is a factor of 2 numbers, it should also be a factor of their sum

Therefore, it should be a factor of 12n+39

If n =2, 7 can be a factor

By hit and trial For no value of n can 21 and 42 become a common factor

Therefore, option A

Answer is C - option . I and II only

We are given:
A=4n−1 and B=8n+40

The difference between B and 2A is:

B−2A=(8n+40)−2(4n−1).
Simplify:
B−2A=8n+40−8n+2=42.
Thus, any common factor of A and B must also divide 42.

The factors of 42 are:
1,2,3,6,7,14,21,42

Check the options.
We are asked to check if
7, 21, and 42 could be common factors of
4n−1 and 8n+40.

Case I: Check 7.
If 7 is a factor of both
4n−1 and
8n+40, then substituting n must satisfy: 4n−1≡0(mod7).

This simplifies to:
4n≡1(mod7).
The modular inverse of 4 modulo 7 is 2, so:
n≡2(mod7)

For 8n+40, substituting

n≡2(mod7) gives:

8(2)+40=16+40=56,which is divisible by 7.
Thus, 7 is a common factor.

Case II: Check 21.

If 21 is a factor of both
4n−1 and 8n+40, then substituting n must satisfy:
4n−1≡0(mod21)

This simplifies to:
4n≡1(mod21).

Solving for
n modulo 21, we find
n≡16(mod21).
Substituting n=16 into 8n+40:

8(16)+40=128+40=168,which is divisible by 21.
Thus, 21 is a common factor.

Case III: Check 42

If 42 is a factor of both 4n−1 and 8n+40, then substituting n must satisfy:

4n−1≡0(mod42).

This simplifies to:
4n≡1(mod42).

Solving for n modulo 42, we find

n≡10(mod42). Substituting n=10 into 8n+40:

8(10)+40=80+40=120
,
which is not divisible by 42.

8(10)+40=80+40=120,which is not divisible by 42.

Thus, 42 is not a common factor.

Final Answer:
The possible factors are 7 and 21.

C. I and II only

​

4n-1 is odd so definitely cannot be divided by an even number (42) but the other two 7&21 can divide both numbers hence choice C

Subtract 4n - 1 from 8n + 40 until n is eliminated

8n + 40 - 4n +1

4n + 41 - 4n +1

= 42

42 is a multiple of 7, 21 and 42

Answer E

How did you deduced so fat that n can be 16? Some trick you applied or you went for each number>

This is helpful. Just want to point out a small typo it's 8n+40, no impact on the solution though.

lets check for 42 first
4n-1 =Even - Odd = Odd , so for a number to be a factor of 4n-1 it has to be odd but all the multiples of 42 are even so here we can eliminate 42

and for lets check rest options
lets simply 8n+40 = 8(n+5) now trick here is to make n+5 as multiple of 7 , 21
let n =2 = 8*7 and 4*2-1 both factors of 7
now lets check 21, 8(n+5) , n= 16 ,8*21 and 4*16-1 = 63 both factors of 21
so the answer is C