<p>For how many positive two-digit integers is the ones digit greater than twice the tens digit?</p>
<p>A) 16
B) 20
C) 28
D) 32
E) 36</p>
<p>--- I want to know the most effective solutions. Counting is very very confusing :(</p>
<p>Really? I think counting is very, very easy.</p>
<p>Make the tens digit a 1. How many two-digit numbers qualify? Not 10, 11, or 12 (because the 2 in the ones place is not strictly greater than twice the 1 in the tens place). But 13, 14, 15, 16, 17, 18 and 19 do. That’s 7 numbers so far.</p>
<p>Make the tens digit a 2. How many two-digit numbers qualify? Not 20, 21, 22, 23 or 24. But 25, 26, 27, 28 and 29 do. That’s 5 more, for a total of 12 so far.</p>
<p>Make the tens digit a 3. How many two-digit numbers qualify? Not 30, 31, 32, 33, 34, 35 or 36. But 37, 38 and 39 do. That’s 3 more, for a total of 15 so far.</p>
<p>Make the tens digit a 4. How many two digit numbers qualify? Not 40, 41, 42, 43, 44, 45, 46, 47 or 48. Only 49 qualifies. That’s 1 more, for a total of 16 so far.</p>
<p>Now make the tens digit a 5. No two-digit numbers will qualify because twice 5 is 10. In order to be greater than twice the tens digit, the ones digit would have to be 11 or more, which is impossible. Same problem occurs when the tens digit is 6, 7, 8 or 9.</p>
<p>So there are 16 numbers that meet the condition stated. As long as you’re systematic about the counting, I don’t think there’s a simpler way to answer this question.</p>
<p>On this one it’s easiest to count</p>
<p>Just use casework based on the tens digit (just like Sikorsky did).</p>