*State Competitions are coming up fast! Are you ready to compete? Let’s try a few 2019 State Competition problems to get ready.*

__2019 State Sprint Round, #18__

If C is a digit such that the product of the three-digit numbers 2C8 and 3C1 is the five-digit number 90C58, what is the value of C?

*Let’s work with just the rightmost two digits. For the units digit, 8 × 1 = 8 does not impact the tens digit. To get the tens digit of the product, we need to cross-multiply the units and tens digits of the two factors: C **× 1 + 8 × C = 9C must end in 5. The only digit for C for which that works is ***C = 5***. *

__2019 State Target Round, #7__

Andy has a cube of edge length 10 cm. He paints the outside of the cube red and then divides the cube into smaller cubes, each of edge length 1 cm. Andy randomly chooses one of the unit cubes and rolls it on a table. If the cube lands so that an unpainted face is on the bottom, touching the table, what is the probability that the entire cube is unpainted? Express your answer as a common fraction.

*When a cube is subdivided along each of the three face-centered axes into n congruent slabs, a block composed of n ^{3} congruent smaller cubes is formed. Each of those smaller cubes has 6 faces, resulting in a total of 6n^{3} faces. Only the outer surface – the 6 faces – of the original cube is painted. Each of the 6 faces of the original cube involves 1 face from each of the n^{2} smaller cubes making up the larger face, yielding 6n^{2} smaller faces that are painted, with the remaining 6n^{3} – 6n^{2} smaller faces unpainted. Removing the outer layer on each face yields an (n – 2) *

*× (n – 2)*

*× (n – 2) cube of totally unpainted smaller blocks, with 6(n – 2)*

^{3}unpainted smaller faces. Thus, with each of the 6n^{3}– 6n^{2}= 6(n^{3}– n^{2}) unpainted smaller faces equally likely, of which 6(n – 2)^{3}correspond to completely unpainted smaller cubes, the probability of landing on a completely unpainted small cube upon landing on an unpainted smaller face is ((n - 2)^{3})/(n^{3}- n^{2})*. When n = 8, the probability is 6*

^{3}/(8^{3}- 8^{2})*= 216/(512 - 64)*

*= 216/448*

*=*

**27/56***.*

__2019 State Team Round, #4__

Suppose that Martians have eight fingers and use a base-eight (octal) number system. If Marty the Martian says he is 37 years old on Mars, how old is he in Earth’s base-ten system?

*Just as 37 as a base-ten number means 3 **× 10 ^{1} + 7 *

*× 10*

^{0}= 3*× 10 + 7*

*× 1 = 37, so 37 as a base-eight number means 3*

*× 8*

^{1}+ 7*× 8*

^{0}= 3*× 8 + 7*

*× 1 = 24 + 7 =*

**31 years***in base ten.*

__2019 State Countdown Round, #12__

For a particular sequence, each term is the sum of the three preceding terms. If *a*, *b*, *c*, *d*, *e*, 0, 1, 2, 3 are consecutive terms of this sequence, what is the value of *a* + *b* + *c* + *d* + *e*?

*As we’re told, each term is the sum of the three preceding terms. In order for this to be true, 2 = 1 + 0 + e, so e must equal 1. Similarly, 1 = 0 + e + d = 0 + 1 + d, so d = 0. Then, 0 = e + d + c = 1 + 0 + c, so c = -1. Continuing this pattern, we find that b = 2 and a = -1. Therefore, a + b + c + d + e = -1 + 2 + -1 + 0 + 1 = ***1***.*

**CHECK THE ****PROBLEM OF THE WEEK ARCHIVE****FOR SOLUTIONS TO PREVIOUS PROBLEMS**