1. ## Consecutive integers (SAT)

Hi,

If $j, k,$ and $n$ are consecutive integers such that $0 < j < k < n$ and the units digit of the product $jn$ is $9$, what is the units digit of $k$?

A. 0
B. 1
C. 2
D. 3
E. 4

I guessed E once but my final answer is C.
The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.

2. Originally Posted by Hellbent
I guessed E once but my final answer is C.
The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.
j,k,n are bigger than zero yes, but might not be 1,2,3 as you have suggested. Instead make them m, m+1 and m+2.

3. Edit: Nevermind, was too slow at typing

4. Originally Posted by pickslides
j,k,n are bigger than zero yes, but might not be 1,2,3 as you have suggested. Instead make them m, m+1 and m+2.
I just realized that only j, k, and n are consecutive.
$jn = 9$

$j = 9$

$n = 11$

$9 \times 11 = 99$

Number intervening 9 and 11 is 10, therefore units digit of k is 0.

5. Also these pairs work

19 21
29 31
39 41
49 51
59 61
... ...

6. This is a problem where you just want to start listing and look for the pattern

1 2 3
2 3 4
3 4 5
4 5 6
......

In less than 30 seconds you should discover that j=9, k=10, n=11 are the first set of numbers that "work."

7. Originally Posted by Hellbent
Hi,

If $j, k,$ and $n$ are consecutive integers such that $0 < j < k < n$ and the units digit of the product $jn$ is $9$, what is the units digit of $k$?

A. 0
B. 1
C. 2
D. 3
E. 4

I guessed E once but my final answer is C.
The question is a bit unclear to me. j, k and n are consecutive such that 0 < j < k < n; so 0 < 1 < 2 < 3, but j times n gives me 1 x 3 = 3 and not 9 as the problem states.
Also, since the integers are m, m+1, and m+2, you can use modular arthimetic.

$m(m+2)\equiv 9 \ \mbox{(mod 10)}$

Therefore, $m = 9+10t \ \ t\in\mathbb{Z} \ \ \ t\geq 0$.

$j=9+10t, \ k=(9+10t)+1, \ n=(9+10t)+2$