<p>If J,K,N are consecutive integer such that 0<J<k<n and the unit(one)digit of the product JN is 9 What is the Unit digit of K?</p>
<ol>
<li> 0</li>
<li> 1</li>
<li> 2</li>
<li> 3</li>
<li> 4</li>
</ol>
<p>I can't do it please answer and explain the solution.</p>
<p>(from math SAT 2008-2009 section 2 number 20)</p>
<p>Thanksssss</p>
<p>I can’t think of a way to explain this quickly and un-awkwardly, but here goes.</p>
<p>The key here is realizing that only the units digits of J, K, and N matter. That’s because you can separate any of these numbers into something that looks like (10x + r), where r is the units digit, and when you multiply them:</p>
<p>(10x + r)(10y + s) = 100xy + 10xs + 10yr + rs</p>
<p>only the ‘rs’ part contributes to the units digit of the product. Everything else goes into another digits place. With that in mind, we want to solve (J)(J+2) = 10a + 9. Add one to both sides:</p>
<p>J^2 + 2J + 1 = (J+1)^2 = K^2 = 10a + 10 = 10(a+1)</p>
<p>So K has to end in zero. Or, alternatively, you could try all the possible values of J (1, 3, 5, 7, and 9) to see that.</p>
<p>I personally really enjoy the explanation above. Interesting approach… Thank you for that contribution.</p>
<p>For this problem, I approached it like this. Since the units digit of JN is 9, we can list the possibilities of J and N. Only 1<em>9(9</em>1) and 3*3 yields 9. J, K and N are consecutive, so use little bit of logic here and conclude that J, K, and N must equal 9, 0, 1 in the units digit, respectively. (ex. 39,40,41 or 49,50,51)</p>
<p>thank guyss, it’s really usuful</p>