IMO Practice Test — Mathematical Reasoning
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
The negation of "Every natural number is greater than 0 and is finite" is:
No natural number is greater than 0 or is finite.
There exists a natural number that is not greater than 0 or is not finite.
Every natural number is not greater than 0 and is not finite.
There exists a natural number that is not greater than 0 and is not finite.
Explanation: Negating "for all (A and B)" gives "there exists (not A or not B)" — De Morgan inside an existential quantifier.
Question 2 of 6
hard
A conditional and its contrapositive always have:
opposite truth values
the same truth value
truth values that depend on p only
truth values that depend on q only
Explanation: p ⇒ q and ~q ⇒ ~p are logically equivalent, so they share the same truth value in every case.
Question 3 of 6
hard
Which statement is a valid disproof of "For every integer n, n² + n + 41 is prime"?
n = 1 gives 43, which is prime.
n = 0 gives 41, which is prime.
n = 40 gives 1681 = 41², which is not prime.
There is no counter-example.
Explanation: n = 40 gives 40² + 40 + 41 = 1600 + 81 = 1681 = 41², which is composite — a genuine counter-example.
Question 4 of 6
hard
If the statement "If it is Sunday, then the shop is closed" is true, which of the following must also be true?
If the shop is closed, then it is Sunday.
If the shop is open, then it is not Sunday.
If it is not Sunday, then the shop is open.
The shop is always closed.
Explanation: Only the contrapositive is guaranteed: "shop is open ⇒ not Sunday" is ~q ⇒ ~p. The others are the converse and inverse, which need not follow.
Question 5 of 6
hard
The statement "(p ∧ q) ⇒ p" is:
a tautology (always true)
always false
true only when p is true
true only when q is true
Explanation: If p ∧ q is true then p must be true; if p ∧ q is false the conditional is vacuously true. So it holds in every case — a tautology.
Question 6 of 6
hard
To prove "if n² is even then n is even", which approach is most natural and why?
Direct method, since assuming n² even directly gives n even.
Contrapositive: prove "n odd ⇒ n² odd", which is straightforward.
A single counter-example.
It cannot be proved.
Explanation: The contrapositive "n odd ⇒ n² odd" is easy (n = 2k+1 gives n² = 2(2k²+2k)+1), and it is equivalent to the original, so it is the natural route.