Unit 2 Cycle 2 Day 27: Iteration + Selection (nested filtering)
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Unit 2, Selection & Iteration • Cycle 2
Day 27 Advanced Practice • Harder Difficulty
Focus: Nested Filtering
Hard
Iteration + Selection (nested filtering)
Advanced Practice Question
Format: Iteration + Selection (nested filtering)
What is printed?
What is printed?
"apcs-keyword">int c = "apcs-number">0;
"apcs-keyword">for ("apcs-keyword">int i = "apcs-number">1; i <= "apcs-number">2; i++)
{
"apcs-keyword">for ("apcs-keyword">int j = "apcs-number">1; j <= "apcs-number">4; j++)
{
"apcs-keyword">if (j % "apcs-number">2 != "apcs-number">0)
{
c++;
}
}
}
System.out.println(c);
Difficulty: Hard |
Topic: Iteration + Selection (nested filtering) |
Cycle: 2 (Advanced)
Why This Answer?
For each i, odd j values are 1 and 3, so c increments 2 times per outer iteration. Two outer iterations → 4.
Common Mistake
Watch Out!
Counting all j values (4 per i) instead of only odd j values (2 per i).
AP Exam Strategy
When a condition filters iterations, count only the values that pass.
Master This Topic
This Cycle 2 HARD question tests nested filtering. Review Unit 2 concepts to build mastery of selection and iteration.
- Understanding nested filtering
- Tracing code execution accurately
- Avoiding common pitfalls
Ready for More Challenges?
Cycle 2 questions prepare you for the hardest AP CSA exam questions.
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