Boston Intellectuals · Mathematics Tournament · Harvard University
Four rounds — Sprint, Target, Team, and Guts — with AMC/AIME-level problems in two divisions (Middle School 6–8 and High School 9–12). Try the samples below to find your level, then register at the bottom of this page.
~30 problems, no calculator, pure speed and accuracy. Short answers only.
Problems in linked pairs (A+B) — your Part A answer feeds Part B. Calculator allowed.
Your team of 4–6 solves together — communication wins. Individuals get assigned a team.
Live-scored relay: run answers to the judges, watch the scoreboard change in real time. Loudest round of the weekend.

Typical: grades 6–7 · school-math comfortable
Sprint-style sample: What is the smallest positive number that leaves remainder 2 when divided by 3, and remainder 3 when divided by 5?
List numbers ≡ 3 (mod 5): 3, 8, 13, 18, 23... Check mod 3: 8 = 3·2 + 2 ✓. Answer: 8. (This exact style appears in our qualification questionnaire — explanation matters more than the answer.)
Team-style sample (from the booklet): A rectangular garden has a perimeter of 56 m and an area of 192 m². What are its dimensions?
Half-perimeter = 28, so length + width = 28 and length × width = 192. Solving: 16 m × 12 m.
Typical: grades 8–9 · AMC 8 / early AMC 10 level
Target-style sample (booklet Pair 2A): Let N be the smallest positive integer with exactly 12 divisors. Find N.
12 = 2·2·3, so exponents+1 multiply to 12. Best assignment to small primes: 2²·3·5 = 60 (divisor count (2+1)(1+1)(1+1) = 12). N = 60.
Linked Part B: using N = 60, what is the sum of all prime factors of N² counted with multiplicity?
60 = 2²·3·5, so 60² = 2⁴·3²·5². Sum with multiplicity = 2·4 + 3·2 + 5·2 = 8 + 6 + 10 = 24. (Booklet lists 25; recompute carefully — this is why Target rewards checking your work: 2+2+2+2 + 3+3 + 5+5 = 24.)
Typical: grades 10–12 · AMC 10/12 – AIME level
Guts-style sample: If x + 1/x = 4, find x³ + 1/x³.
Cube the identity: (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x). So 64 = x³ + 1/x³ + 12, giving x³ + 1/x³ = 52.
Target-style sample (booklet Problem 10): For how many integers n with 1 ≤ n ≤ 100 is n³ − n² divisible by 4?
n³ − n² = n²(n − 1). Check n mod 4: it's divisible by 4 whenever n ≡ 0 or 1 (mod 4), and also when n is even (n² carries a factor of 4). Counting all valid n from 1–100 gives 75.
Olympiad Level · ★ · Geometry + Trigonometry
In triangle ABC, the angle bisector from vertex A meets BC at point D. The circumradius is R = 7, the inradius is r = 3, and the angle at A is 60°. Find the length AD in exact form. This is a real Target Bonus problem from the Freedom Math Tournament booklet.
AD = 2r / sin(A/2) = 6 / sin(30°) = 6 / (1/2) = 12. The key identity links the inradius, the half-angle at A, and the distance along the bisector. From the booklet's Target Bonus.
Linked Part B: D divides BC in ratio BD:DC = 2:3 with BC = 10, so BD = 4. With R = 7, r = 3, A = 60°, the area of triangle ABC = 21√3, and Area of triangle ABD = (2/5)·21√3 = 42√3 / 5.
You compete only against your own division (MS 6–8 or HS 9–12). Individual score = Sprint + Target; team awards combine Team + Guts. Awards: gold, silver, and bronze medals, trophies, and cash prizes for champions — every contestant receives a certificate. No prior competition experience is required, and homeschooled students register as independent.
Found your division and level?
Register below to secure your seat.▼