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

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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)

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

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