12 Days of Christmas GMAT Competition - Day 2: If n is an integer : Problem Solving (PS)

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17 Dec 2024, 10:11

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4n-1 is an odd integer & 8n+40=8(n+5) is even

if n=2.. 7 & 8*7, both are divisible by 7.
if n=16, ..63 & 8*21 , both divisible by 21

42 = 2*3*7..can only divide something even. Hence 4n-1 is not divisible by 42.

Thus 7 & 21.

Answer C

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17 Dec 2024, 10:18

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Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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Let's see for 7,
For n=2, 4n-1 and 8n+40 both will be divisible by 7.

For 21,
8n+40 = 8(n+5) should be divisible by 21
Now 8 isn't divisible by 21 so n+5 must be divisible by 21,
We can check n+5= 21,
n=16.
Yes, for n=16, 8n+40= 168 which is divisible by 21.
Similarly for n= 16, 4n-1 is also divisible by 21.

For 42,
4n-1 will never be divisible by 42 as 4n-1 will always yield an even value and to be divisible by 42, the number has to be even.

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17 Dec 2024, 10:25

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Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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The difference between 4n-1 and 8n+40 could be a factor to resolve the question.
4n-1 to make it comparable, we double it and reach 8n - 2.

8n+40 - (8n-2)
= 42

Either 7, 21 or 42 could be a factor of 42, therefore the answer is E

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17 Dec 2024, 10:35

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If n=9, 4n-1 and 8n+40 are divisible by 7.
If n=16, 4n-1 and 8n+40 are divisible by 21.
Tried some numbers but they're not divisible by 42
I'm not sure that I tried the exact method.

Option C

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17 Dec 2024, 10:37

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For 4n−14n - 14n−1 and 8(n+5)8(n + 5)8(n+5),
checking values of n

4n−1 = 3, 7, 11, 15, 19, 23, 27,....

only 7 is present in this sequence
so Option 1

Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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17 Dec 2024, 10:40

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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

If n =2 then each of the terms are divisible by 7 and if n=16 then each of the terms is divisible by 21. Since 4n-1 is odd hence it is not divisible by 42.

We can also see difference between both terms and evaluate.

Ans C

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17 Dec 2024, 11:00

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So

Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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So if 4n-1 = 7k, 4n = 7k+1,
8n+40 = 14k + 42. This is also divisible by 7.

Similarly, 42k+42, 84k+42... So both 21,42 are also factors.

Hence I,II,III is the correct choice.

IMO E

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17 Dec 2024, 11:03

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4n-1 and 8n+40= 8(n+5)
n=2 - 7 will be the factor of both.
for 21= 7*3 put n=16 as 16+5 = 21- 21 will be the factor of both.
for 42= 7*3*2= for this remember 4n-1 will be odd and 42 is even o/e doesnt happen so it will never be the factor.

ANS- C

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17 Dec 2024, 11:06

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Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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factor of 4n-1 & 8n+40 ==> 8(n+5)
substitute for 7 to be a factor of 8(n+5) n=2,9,... ==> 4n-1==> 4*2-1==>7 which is also a factor of 7
similarly for 21 to be a factor n=16
42=2*21 ==> n=16

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17 Dec 2024, 11:15

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IMO C

So just by seeing can elimante option 3 ie 42 as 4n-1 can not be a factor of 42 reason being it is always odd and multiple of 42 is always even.

Now for 1 and 2 can go by try and error and have values n=2 and n=16

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17 Dec 2024, 11:55

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Maybe there was a better way and some number properties to be applied but I used random numbers to come to the ans - C both 1 and 2 i.e. 7 and 21 are valid

7 is valid when n = 2
21 is valid when n = 16

I took some time for the 3rd case where I realised that 42 is even and any multiple will be even. We have 4n - 1 = even -> 4n = even + 1 = Odd
so n cannot be an integer and hence 42 cannot be a factor of this. So C.

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Factor of both 4n-1 and 8n+40.
The sequence for 4n-1 is as follows - 3,7,11,15,19,23,27,31,35,39,43,47,51,55,59,63...... (unit digit ending in 3,7,1,5,9)
The sequence for 8n+40 - 48,56,64,72,80,88..... (unit digit ending in 0,2,4,6,8)
Out of 7,21 and 42 - 7 cannot be a factor because in the second sequence, there is no multiple of 7
42 cannot be a factor as in the first sequence there is not multiple of 42
21 can be a factor since its multiples have unit digits from 1 to 9

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17 Dec 2024, 12:06

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Playing with numbers we see that choosing n=2 the values of the expressions are 7 and 56, both divisible by 7.
The same happens with n=16, 63 and 168 are divisible by 21.
As 4n-1 is always odd, it can not be divisible by 42.

IMO C

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Let c be a factor of the expressions.

If c | (4n - 1) et c | 8n + 40, when c |(8n + 40)v - (4n - 1)u , where u and v integers.
Let’s take v = 1 and u = 2.
Then, (8n + 40)v - (4n - 1)u = 8n + 40 - 8n + 2 = 42.

7, 21 and 42 are all factors of 42.

Answer: E

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17 Dec 2024, 12:12

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Answer is C. I and II only

4n - 1 must be an odd integer, and 8n + 40 must be a multiple of 8.

I. 7 could be a factor; for n = 2, 4n - 1 = 7 and 8n + 40 = 56.
II. 21 could be a factor; for n = 16, 4n - 1 = 63 and 8n + 40 = 168.
III. 42 could not be a factor; it is even and any factor multiplied by an even factor is even, which would not satisfy 4n - 1.

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17 Dec 2024, 12:30

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Given two integers are (4n-1) and (8n+40)

Rearranging second one
8n+40 = 8n - 2 + 2 + 40 = 2(4n-1) + 42

For a number to be factor for both integers - both numbers need to be divisible by the factor

Therefore any possible option should be a factor of 42 as well. All three options satisfy this

Checking some possible numbers for options
1. When n = 2, First integer is 7 -> so 7 is a possible factor
2. When n = 16, First integer is 63 -> so 21 is a possible factor
3. 4n-1 is always odd, therefore no even number(42) cannot be factor of first integer

1 and 2 are valid (OPTION C)

Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

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we need to check if any of these COULD be a common factor of both 4n-1 and 8n+40.

8n+40 can be written as 8(n+5).

The way i approached this is, since the factor of 8(n+5) will be all the factors of 8 and all the factors of (n+5).

For 7, since 8 is not a multiple of 7, make n+5 = 7.
that gives n = 2.

Put n=2 in both,

4n-1 gives 7.
8(n+5) gives 8*7.

Hence, 7 is a factor.

For 21, since 8 is not a multiple of 21, make n+5 = 21.
This gives n = 16.

Put n=16 in both,

4n-1 gives 63.
8(n+5) gives 8*21.

Hence, 21 is a factor.

For 42, since 8 is not a multiple of 42, make n+5 = 42.
this gives n = 37

put n = 37 in both,

4n-1 gives 147, which is not a multiple of 42, hence not a factor.

Therefore,

Final answer, only 1 and 2 are factors. OPTION C

Bunuel

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

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