# Consecutive integers (SAT)

• Jan 13th 2011, 12:59 PM
Hellbent
Consecutive integers (SAT)
Hi,

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

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.
• Jan 13th 2011, 01:04 PM
pickslides
Quote:

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.
• Jan 13th 2011, 01:11 PM
Quacky
Edit: Nevermind, was too slow at typing (Worried)
• Jan 13th 2011, 03:51 PM
Hellbent
Quote:

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.
$\displaystyle jn = 9$

$\displaystyle j = 9$

$\displaystyle n = 11$

$\displaystyle 9 \times 11 = 99$

Number intervening 9 and 11 is 10, therefore units digit of k is 0.
• Jan 13th 2011, 03:58 PM
pickslides
Also these pairs work

19 21
29 31
39 41
49 51
59 61
... ...
• Jan 13th 2011, 05:33 PM
DrSteve
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."
• Jan 13th 2011, 05:48 PM
dwsmith
Quote:

Originally Posted by Hellbent
Hi,

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

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.

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

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

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