<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>I used the guess and check method but it didn't work out too tell ... killed my brain. I need your help guys</p>
<p>I think it is zero. For the product to end in a 9, I think you need end digits 3x3 or 1x9 or 7x7, so I think J and N end in 9 and 1 and the three consecutive integers are 9, 10, 11 or 29, 30, 31, something like that.</p>
<p>its 0 b.c jn are apart by 2(n is 2+j), so just multiply combinations like 1x3 and 2x4 so on until you get a product which has 9 in its units place. For example 9x11=99 so j k and n could be 9, 10 , and 11.</p>
<p>for 9 to be in the units digit, j and n’s last digits must be either 3,3 or 1,9 or 4,4
Since j<n is mentioned
j and n must have 1 and 9 as units digits in either way.
So we could have j,k,n as 9, 10, 11 or 29, 30, 31 etc.
so k must end in 0.</p>
<p>is this from the may sat?</p>
<p>The answers that supports the choices you gave us is that K ends with 0 or 3. The integers can be 9,10,11 or 109,110,111 (on a bigger scale) if k=0 . If k=3 then you could use 19/20/21 and combination like that.</p>