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

Question

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

If n is an integer, which of the following could be a factor of both 4n - 1 and 8n + 40?

I. 7 
II. 21
III. 42

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

First, notice that since n is an integer, 4n - 1 will always be odd. This means it cannot have any even factors, so 42 is eliminated from the start.

Next, let's rewrite 8n + 40 so that 4n - 1 appears in the expression. This could help us deduce a connection between the two numbers:

8n + 40 =

= (8n - 2) + 42 =

= 2(4n - 1) + 42.

Now, we need to find a divisor, d, such that both (4n - 1)/d and (2(4n - 1) + 42)/d are integers. Observe that:

\frac{2(4n - 1) + 42}{d}= \frac{2(4n - 1)}{d} + \frac{42}{d } .

Now, if (4n - 1)/d is an integer, then 2(4n - 1)/d will also be an integer. This means that for   \frac{2(4n - 1)}{d} + \frac{42}{d }   to be an integer, 42/d must also be an integer. Consequently, d must be an odd factor of 42.

The odd factors of 42 are 1, 3, 7, and 21 (and their negatives).

Answer: C. I and II only.

YashK23 · Answer

Option C is the Ans.

For n = 2, we get 7 and 56, which are both divisible by 7. For n = 16, we get 63 and 168 which are both divisible by 21. For any value, 4n - 1 will not yield an even number, so 42 can never be a factor of 4n - 1. Hence option C is correct.

Krunaal · Answer

For 7 to be a common factor, let's check when  n = 2 ; as it gives 7 when substituted in 4n-1. When substituted in 8n-40, it gives 56 => which has 7 as a factor.  Therefore, 7 can be a factor of both.

For 21 to be a common factor, let's check when  n = 16 ; as it gives 63 when substituted in 4n-1 and 63 = 21*3. When substituted in 8n-40, it gives 168 => 21*8.  Therefore, 21 can be a factor of both.

For 42 to be a factor of both, each should be an even number as 42 = 21*2, but we can see that  4n-1 will always be odd .  Therefore, 42 cannot be a factor of both.

Answer C.

hr1212 · Answer

We are given two equations - 4n-1 and 8(n + 5) and we need to calculate its factors from given options for any integer n.

Before we start, let's closely look at both these numbers. By initial analysis, we can confidently say that first number is always an odd number whereas second number is always an even number.

Now, by this analysis we can conclude that the first number cannot have an even factor.

Let's analyze all the options now,

n 
 4n-1 
 8(n+5) 
 
  2 
 7 
 8*7 
 
  16 
 63 = 3 * 21 
 8*21 
 
  ?? 
 42k != odd 
 4 * 42

As, we cannot find an integer n for 4n - 1 such that it has 42 as one of it's factor, whereas we can easily deduce n for values 7 and 21, so the correct answer is (I) and (II)

Answer : C

Kinshook · Answer

If n is an integer, which of the following could be a factor of both 4n - 1 and 8n + 40?

If n=1; 4n-1=3; 8n+40 = 48; 3 is a factor of both 4n-1 & 8n+40
If n=2; 4n-1=7; 8n+40 = 56; 7 is a factor of both 4n-1 & 8n+40
If n=3; 4n-1=11; 8n+40 = 64; No common factor of both 4n-1 & 8n+40
If n=4; 4n-1=15; 8n+40 = 72; No common factor of both 4n-1 & 8n+40
If n=5; 4n-1=19; 8n+40 = 80; No common factor of both 4n-1 & 8n+40
If n=6; 4n-1=23; 8n+40 = 88; No common factor of both 4n-1 & 8n+40
If n=7; 4n-1=27; 8n+40 = 96; 3 is a factor of both 4n-1 & 8n+40
If n=8; 4n-1=31; 8n+40 = 104; No common factor of both 4n-1 & 8n+40
If n=9; 4n-1=35; 8n+40 = 112; No common factor of both 4n-1 & 8n+40
If n=10; 4n-1=39; 8n+40 = 120; No common factor of both 4n-1 & 8n+40
If n=11; 4n-1=43; 8n+40 = 128; No common factor of both 4n-1 & 8n+40
...
f n=16; 4n-1=63; 8n+40 = 168; 21 is a factor of both 4n-1 & 8n+40
.....
42 can not be a common factor of both 4n-1 & 8n+40 since any multiple of 42 is even and 4n-1 is always odd.

3, 7 & 21 are common factor of both 4n-1 & 8n+40

I. 7 
II. 21
III. 42

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

IMO C

Mantrix · Answer

Two Expressions we have

1. 8(n+5)
2. 4n-1

So, we will check with the option, as the given options can be the factors of given option or not?

1. 7

Take n =2, it will give me 1. 56, 2. 7, so yes we can see that 7 is the factor of both the expressions

2. 21  
Take n=16(Predicted by n+5=21), it will give me 1. 8(n+5) = 8*21, 2. 4n-1 =63, yes we can see that 21 is the factor of both the expressions

3. 42 
Take n = anything, it will not satisfy 4n-1/42 (as 42 is the even number and numerator is the odd)

so from here we got only I and II

dtxgmat · Answer

OPTION C

Using exclusion remove option (III), as you can see that 4n - 1 will always be ODD (EVEN - ODD = ODD) 
 42 cannot be a factor to an odd number

Evaluating Option I.  
 
 Assume: 7 is a factor of 4n-1

= 7k = 4n-1 (where k is an integer) 
 = n =  \frac{7k+1}{4}  --------- (Eq. 1)

To prove, that 8n + 40 is also divisible by 7, dividing them both should leave no remainder

Evaluating, 
 = 8n + 40

replace "n" from Eq 1

= 8 * ( \frac{7k + 1}{4} ) + 40 
 = 2 * (7k + 1) + 40 
 = 14k + 2 + 40 
 = 14k + 42 
 = 7 (2k + 6) 
 this is a multiple of 7, as we assumed k as an integer

Hence, 7 is factor of 8n + 40 too

Solve similarly for 21, which will come as true.

juliansorre · Answer

This question is harder than it seems.

You can always solve it by picking up some values, and if you are lucky, you can solve it rapidly. However, a more comprehensive solution requires the knowledge of advanced Number Theory, which I don ́t beleive is part of the GMAT curriculum.
This question, in particular, involves the Congruence Theory.

We write:

a≡b(mod m)

and this means:

a and b leave the  same remainder  when divided by m.

Equivalently:

m∣(a−b)

which means  m divides a−b .

We want:

m∣(4n−1) for any m odd greater than or equal to 3.

This is equivalent to:
4n−1≡0(mod m)
Rearranging:
4n≡1 (mod m)

General Theorem:

If gcd⁡(a,m)=1, then the equation:
ax≡1(mod m)
always has a solution.

In case we have a = 4, and m is every odd integer greater than or equal to 3.

Clearly, gcd(4,m) =1

so 4n -1 is always divisible by an integer greater than or equal to 3.

Sunsh · Answer

4n -1 ..... 4n is always Even, So 4n-1 will be Odd
8n+40.... 8n always Even, 8n+40 is Even
 Now.. if we factorize all the numbers.. 7 Odd
 21 = 3 x 7 ; Both Odd
 42 = 2 x 3 x 7 ; Even..
Now Odd number cannot have Even Factor, So 42 is not possible.

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

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

Prep Toolkit

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