If n and k are positive integers and n + k

Question

If n and k are positive integers and n + k < 8, how many different values of the product nk are possible

A. 8
B. 9
C. 10
D. 12
E. 16

egmat · Accepted Answer

Watch this video solution to see how you can solve this deceptively simple question without making any inadvertent errors.
 https://www.youtube.com/watch?v=PTZ2qYmZi5o

The key - Making a grid to visualize all the pertinent combinations so that you do not miss out on any.

gmatophobia · Answer

We can count the possibilities as shown below. Total number of values = 10 Screenshot 2023-10-29 101952.png Option C

chetan2u · Answer

So n+k can be any integer from 2 to 7, since n+k is at least 2 and max 7.

let one of them be 1, then the second could be anything from 1 to 6.
Product = 1, 2, 3, 4, 5, 6

Next, let one of them be 2, then the second could be anything from 1 to 5.
Product = 2, 4, 6, 8, 10

Now, let one of them be 3, then the second could be anything from 1 to 4.
Product = 3, 6, 9, 12

All further values will be repetition as values of n and k will interchange but will no affect the value of nk.

Distinct values = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, that is 10 values

C

edwin.que · Answer

My confusion is that if n and k are represented in different letters, so that they cannot be the same number. For example, n=1 and k=1 would not be possible.

gmatophobia · Answer

That's not a correct inference  edwin.que . The question doesn't say that n and k are distinct integers. Hence, we can have a possibility that n and k have the same value.

Jahnavi_M · Answer

Maximum of nk is when n = k
=> 2 <= n + k < 8
=> 2 <= 2 * n < 8
=> 1 <= n^2 < 16
Hence, max is 15 and min is 1

1 - 1 * 1 ( 1 )
2, 3, 5 -> all 1 * x form and 1 + x < 8 is true for all of them (4)
Other primes: 7, 11, 13 => 1 + x >= 8

Now,
4 -> 1 * 4 (5)
6 -> 1 * 6 (6)
8 -> 1 * 8 not possible but 2 * 4 is possible (7)
9 -> 1 * 9 not possible but 3 * 3 is possible (8)
10 -> 1 * 10 not possible but 2 * 5 is possible (9)
12 -> 1 * 10 no, 2 * 6 no 3 * 4 is possible (10)
14 -> 1 * 14 no, 2 * 7 as well no
15 -> 1 * 15 no, 3 * 5 as well no.

So, total = 10

Posted from my mobile device

sachi-in · Answer

n + k < 8

now from AM >= GM always and atmost AM = GM
so \(n + k >= 2 * \sqrt{nk}\)

Highest value of geometric mean ( when GM = AM ) is when N = K or N as near to K as possible

Here max of n * k will be when n nearly equal to k ,
since n + k < 8
the (n, k) = (3, 4) or (4, 3) pair will give us the highest value for n*k = 12

Now max ( n * k ) = 12 also n and k are positive integers so none are zero
Hence All composite numbers between 1 to 12 whose both factors are less than 8 are possible as multiples of n * k

1 2 3 4 5 6 8 9 10 12 ( 7 = 7 * 1 and 11 = 11 * 1 which are not possible )

samarpan.g28 · Answer

n and k are positive integers. n+k<8 i.e. the sum can be from 1 to 7. We can consider the below values for n and k.
n    k                 products
1    1,2,3,4,5,6   1,2,3,4,5,6
2    1,2,3,4,5      2,4,6,8,10
3    1,2,3,4         3,6,9,12
4    1,2,3            4,8,12
5    1,2               5,10
6    1                  6
unique values are 1,2,3,4,5,6,8,9,10,12 i.e. 10 values. Option (C) is correct.

Kinshook · Answer

edwin.que 
If n and k are positive integers and n + k < 8, how many different values of the product nk are possible
n + k < 8
(n, k) = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),​​​​​​​​(5,1),(5,2),(6,1)}

nk = {1,2,3,4,5,6,8,9,10,12}: 10 different values of nk

​​​​​​​IMO C

shwetakoshija · Answer

Given : n + k < 8 for positive integers n and k.

So, n + k is at most 7.
Simply, start writing possible pairs of n and k by taking different possibilities for n:

(1)  n = 1 : Then, (n, k) can be (1, 1), (1, 2), ..., (1, 6).
 
 There are  6 possible products  (nk) here: 1, 2, ..., 6

(2)  n = 2 : This time, we will not write (2, 1) since this combination of n and k has already been considered in Case 1  . Note that in a product, order of n and k does NOT matter.
 
 So, new pairs are: (2, 2), (2, 3), ..., (2, 5). 
 The products we get here are: 4, 6, 8, 10. 
 Since 4 and 6 are already accounted for in Case 1, only  two new products  come from here: 8 and 10.

(3)  n = 3 : Again, new pairs are: (3, 3), (3, 4)
 
  Two new products  from here are 9 and 12.

We can stop here since no more new pairs of n and k can be created given the constraints in the question.

Hence, total number of possible values of nk are 6 + 2 + 2 =  10 .

DanTheGMATMan · Answer

Realistically, it's probably best just to count out possibilities and just think of what products are possible within the constraints, in an organized way:

https://www.youtube.com/watch?v=bLo76sTH0CI

KarishmaB · Answer

One method is to enumerate as done above.

Another is to realize that the sum of n and k can be at most 7 and values of n and k vary from 1 to 6. 
This means that the product will be maximum when n = k = 3.5. But since they must be integers, product will be maximum when they are as close as possible i.e. 3 and 4.
So maximum product is 3*4 = 12

Now it is simple to check whether all 12 products are possible. 
If n = 1, k can be anything from 1 to 6. So first 6 products are possible.

Product of 7 is not possible because neither n nor k can be 7. 
Product of 8 (= 4*2), 9 (3*3), 10 (2*5) and 12 (3*4) are possible. 
Product cannot be 11 because neither n nor k can be 11.

Hence we get 10 possible distinct values of nk.

Answer (C)

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

If n and k are positive integers and n + k < 8, how many different

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions