Is it just me or is this not-yet-doable....

Test Preparation SAT Preparation
1

<ol>
<li>How many positive four-digit even integers have a 5 in
the thousands place, a 2 in the tens place, and at least
one of the digits equal to 6? </li>
</ol>

<p>(A) 10
(B) 12
(C) 14
(D) 16
(E) 18</p>

2

<p>OK, you have two ways to make this happen: if you put a 6 in the ones place, you can put any digit (0 - 9) in the 100s place. That gives you 10 numbers. Or you can put the 6 in the 100s place and then you have to put 0, 2, 4, 6, or 8 in the ones place. That’s 5 more numbers. But wait: we already counted 66 in the first group! </p>

<p>So the final answer is 10 + 5 -1 = 14.</p>

<p>Tricky question. Where’s it from?</p>

3

<p>Is it C.14?</p>

4

<p>its from an actual PSAT and yea its C. Now I know why I got it wrong…forgot to -1</p>

<p>thanks</p>

5

<p>Where do you get actual PSAT’s?</p>

6

<p>reduce it to a 2 digit integer:</p>

<p>_ _ </p>

<p>The first digit can be 0,1,2,3,4,5,6,7,8,9
last digit can be 0,2,4,6,8</p>

<p>assume that the first is the six, then you have 5 options</p>

<p>assume the first isn’t a six, then you have 9 options</p>

<p>Thus total is 14 ===> C</p>

7

<p>i didnt get it at first but this helps</p>

<p>there’s only 1 number possible to have a 5 in the thousands place (5000). a 2 in the tens place: only 1 possible (5020). since there can be 10 digits in the hundreds place (5020, 5120, 5220, 5320…5920). only possible combo for a number for one of the digits to equal 6 are: 5620, and 5626.
1+1+10+1+1=14</p>