<p>If j,k, and n are consecutive integers such that 0<j<k<n and 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>If the answer is A, then this is how I did it:</p>
<p>0<j<k<n So if j*n (their one’s digits) = 9 Then j or n have to end in either 1 and 9 or 3 and 3 (e.g. 0<29<k<31) and no consecutive integers can fill the case for 3 and 3. Therefore k would have to end in 0.</p>
<p>Oh… wow! I only thought of 3,3 and totally disregarded 9,1. Thanks for the help! (This was a lvl5 question btw)</p>
<p>Let k = n-1 and j = n-2</p>
<p>We have: n(n-1)(n-2) </p>
<p>With n(n-2) = x9 So n(n-2) = 09,19,29,39 etc…= 9 + 10n, n>0</p>
<p>n = 11, n-2 = 9, n(n-1) = 99</p>
<p>So k = n-1 = 10 Units = 0</p>
<p>answer (a)</p>
<p>(0<9<10<11)</p>