Consider this C program (preserve the exact newlines):\n#include<stdio.h>\nint main()\n{\n int x = 10, y = 100%90, i;\n for(i = 1; i < 10; i++)\n if(x != y);\n printf("x = %d y = %d\n", x, y);\n return 0;\n}\n\nWhich statements about its output and syntax are correct?

Introduction / Context:
This problem examines how a stray semicolon after an if statement and the absence of braces affect control flow, as well as the behavior of the modulus operator.

Given Data / Assumptions:

• x is 10; y is 100 % 90, which equals 10.
• for loop iterates i from 1 to 9.
• There is a semicolon immediately after if(x != y).
• printf is indented but not braced beneath either if or for.

Concept / Approach:
Because of the trailing semicolon, if(x != y); becomes a null statement that executes every loop iteration. The printf line is not part of the if and, due to lack of braces, it is also not part of the for; it executes once after the loop completes. Since x and y are both 10, the printed values are x = 10 y = 10.

Step-by-Step Solution:
Compute y: 100 % 90 → 10.Loop: for (i = 1; i < 10; i++) executes a useless if; nothing else in the loop body.After the loop ends, run printf once and print “x = 10 y = 10”.

Verification / Alternative check:
Add braces around printf to bind it to for and observe 9 prints; remove the semicolon to bind it to the if and see conditional printing.

Why Other Options Are Wrong:
“1 only” claims 10 prints, which is false here. “3 and 4” wrongly asserts no output. “4 only” also says no output, which is incorrect. “1 and 2” includes a false statement.

Common Pitfalls:
Relying on indentation rather than braces; missing the effect of a stray semicolon after if; misunderstanding operator % and precedence.

Final Answer:
2 and 3