Grade 7 Math Olympiad Problems

Problem 1: a sequence built from its own past terms

A sequence starts 3,53, 5, and every term after that is the sum of the previous two. Find the 8th term.

Solution: Just build it term by term:

3,  5,  8,  13,  21,  34,  55,  893,\; 5,\; 8,\; 13,\; 21,\; 34,\; 55,\; 89

Each term is the sum of the two before it (e.g. 21+34=5521+34=55), so the 8th term is 89.

Problem 2: matching bases in disguise

If 2n×3n=2162^n \times 3^n = 216, find nn.

Solution: Combine the two powers first, since they share an exponent:

2n×3n=(2×3)n=6n2^n \times 3^n = (2\times3)^n = 6^n

So 6n=216=636^n = 216 = 6^3, giving n=3n=3.

Problem 3: two overlapping groups

In a class, 60% of students play basketball, 50% play soccer, and 30% play both. What percentage play neither?

Solution:Add the two sports, then subtract the overlap once so it isn't double-counted:

Play at least one=60%+50%−30%=80%\text{Play at least one} = 60\% + 50\% - 30\% = 80\%

So 100%−80%=20%100\% - 80\% = 20\% play neither sport.

Problem 4: three consecutive integers

Three consecutive integers add up to 333. Find the largest one.

Solution: Call the smallest nn, so the three are n,n+1,n+2n, n+1, n+2:

n+(n+1)+(n+2)=333  ⇒  3n+3=333  ⇒  n=110n + (n+1) + (n+2) = 333 \;\Rightarrow\; 3n+3=333 \;\Rightarrow\; n=110

The three integers are 110, 111, 112. The largest is 112.

Problem 5: a clock that wraps around

A 12-hour clock shows 3:00. What will it show 100 hours later?

Solution: Only the remainder after dividing by 12 matters, since every full 12 hours brings the clock back to the same position:

100=8×12+4⇒remainder 4100 = 8 \times 12 + 4 \quad \Rightarrow \quad \text{remainder } 4

Adding just the remaining 4 hours to 3:00 gives 7:00.