Puzzle:

Find a cycle of five 4-digit numbers such that the last 2 digits of each number are equal to the first 2 digits of the next number in the cycle. Each of the 5 numbers must be exactly one of the following types (with each of the 5 types being represented exactly once): Square, Cube, Triangular, Prime, Fibonacci. The solution is unique.

Solution:

8117, 1728, 2850, 5041, 4181 (Prime, Cube, Triangular, Square, Fibonacci)

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Solution:

8117, 1728, 2850, 5041, 4181 (Prime, Cube, Triangular, Square, Fibonacci)

## 4 comments:

5041 square(71^2)

1728 cube(12^3)

8117 prime( no factors)

4181 Fibonacci(1 1 2 3 5,8,13,21,34,55,89,144,233,377, 610,987, 1597, 2584, 4181)

2850 triangular( 75 the triangular number)

2850, 5041, 4181, 8117,1728

2850 is 75th triangular number

Amiable brief and this enter helped me alot in my college assignement. Thank you on your information.

Opulently I agree but I think the brief should have more info then it has.

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