<p>If j, k, and n are consecutive integers such that 0 < j < k < n and the units (ones) digit of the product jn is 9, what is the units digit of k?</p>
<p>(A) 0
(B) 1
(C) 2
(D) 3
(E) 4</p>
<p>How would you solve this problem. I got the correct answer (A), but it took a lot of guess-and-checking-based-on-reason. How would you solve this algebraically?</p>
<p>There are 2 options for the units digits of j and n. 3 and 3, or 9 and 1. The only possible one for a consecutive integer would be 9,0,1.</p>
<p>I don't think there's an algebraic way to solve it, or at least I don't think there's an easy algebraic method.
Since the units digits of j and n multiply to something ending in 9, the units digits of j and n must be either 3,3; 1,9; and 9,1. It can't be 3,3 because it's consecutive, and it can't be 1,9 because that would not allow for k that is consecutive with j and n. That leaves 9,1 as the only option, where the units digit of k is 0.</p>
<p>here's how i solved it : okay, so jn = X9 where X = 1,2,3,4,5,6,7,8,9,0.... now which of those digits is a good number that is divisible by 2 other numbers... not 19, not 29, not 39, not 49 (cuz both j and n cant be equal).. not 59, not 69, not 79, not 89, 99!!! 99 = 11 x 9.... or 9 X 11.. so if j = 9.. and n = 11... the number between them = 10! ending with a 0.</p>
<p>^cool,keeprolling</p>