The 82nd William Lowell Putnam Mathematical Competition, 2021

William Lowell


maa.org/putnam
PUTNAM
Mathematical Competition
Problems for
Session A

The 82nd William Lowell Putnam Mathematical Competition
2021

, William Lowell


maa.org/putnam
PUTNAM
Mathematical Competition
Problems for
Session B

The 82nd William Lowell Putnam Mathematical Competition
2021

, A1 A grasshopper starts at the origin in the coordinate plane and makes a sequence of hops. Each
hop has length 5, and after each hop the grasshopper is at a point whose coordinates are both
integers; thus, there are 12 possible locations for the grasshopper after the first hop. What is
the smallest number of hops needed for the grasshopper to reach the point (2021, 2021)?
Answer: 578.
Solution: Each hop can be described by a displacement vector hp, qi with p2 + q 2 = 25 ; the
twelve possible vectors are

h3, 4i; h−3, 4i; h3, −4i; h−3, −4i; h4, 3i; h−4, 3i; h4, −3i; h−4, −3i; h5, 0i; h−5, 0i; h0, 5i; h0, −5i.

One way to write the total displacement as a sum of 578 of these vectors is

h2021, 2021i = 288 · h3, 4i + 288 · h4, 3i + h0, 5i + h5, 0i.

To show that it cannot be done with fewer, note that each hop can increase the sum of the
grasshopper’s coordinates by at most 3 + 4 = 7. Because this sum has to reach

2021 + 2021 = 4042 = 7 · (577) + 3 ,

at least 578 hops are needed.




1