February 27, 2017

State competitions officially start this week! Are you ready to compete? Let’s warm up with a few problems from the 2016 State Competition.

The digits of a 3-digit integer are reversed to form a new integer of greater value. The product of this new integer and the original integer is 91,567. What is the new integer?

[Sprint #19]       

Since the last digit of our product is a 7, let’s look at multiplying 7 × 1. Consider the product shown. Looking at the far left column, the ten-thousands place, the product of our two numbers has a 9 here but we see a 7 above it. This tells us that 2 must be added, which will be carried over from the thousands column. So B + 7B + N = 21, where N represents a single digit integer number carried over from the hundreds column. Rearranging we get  8B + N = 21. If B is 1, N would be 13; if B is 2, N would be 5; and if B is 3, N would be –3. The only way to get N to be a single digit positive integer is if B = 2. This means our multiplication would be 127 × 721 = 91567. Therefore, the new integer is 721.



A spinner is divided into 5 sectors as shown. Each of the central angles of sectors 1 through 3 measures 60° while each of the central angles of sectors 4 through 5 measures 90°. If the spinner is spun twice, what is the probability that at least one spin lands on an even number? Express your answer as a common fraction.

[Sprint #21]        

There are two even numbers, 2 and 4. The section labeled 2 is 60 degrees, and the section labeled 4 is 90 degrees. The probability of one spin landing on an even number is, therefore, (60 + 90)/360 = 150/360 = 5/12. The probability of one spin landing on an odd number is 1 − 5/12 = 7/12. If you are spinning twice, there are three ways to get at least one even. You can spin two evens, an even and then and odd or an odd and then an even. The probability of spinning two evens is 5/12 × 5/12 = 25/144. The probability of spinning an even then an odd is 5/12 × 7/12 = 35/144. The probability of spinning an odd then an even is 7/12 × 5/12 = 35/144. So the probability of spinning at least one even is 25/144 + 35/144 + 35/144 = 95/144.



In the list of numbers 1, 2, …, 9999, the digits 0 through 9 are replaced with the letters A through J, respectively. For example, the number 501 is replaced by the string “FAB” and 8243 is replaced by the string “ICED”. The resulting list of 9999 strings is sorted alphabetically. How many strings appear before “CHAI” in this list?

[Sprint #29]        

Alphabetic sorting is just what you find in the dictionary. Let’s consider the cases of 1-letter, 2-letter, 3-letter and 4-letter strings represented by 1-digit, 2-digit, 3-digit and 4-digit numbers, respectively.

1-digit numbers: The 1-digit numbers 1 and 2 represent B and C, which come before CHAI alphabetically. That’s 2 strings.

2-digit numbers: The 2-digit numbers from 10 through 19 and from 20 through 27represent the strings BA, BB, BC, …, BJ and CA, CB, CC, …, CH, respectively, which all come before CHAI alphabetically. That’s a total of 9 + 9 = 18 strings.

3-digit numbers: The 3-digit numbers from 100 through 199 and from 200 through 270 represent the strings BAA, BAB, BAC, …, BJJ and CAA, CAB, CAC, …, CHA, respectively, which all come before CHAI alphabetically. That’s a total of 100 + 71 = 171 strings.

4-digit numbers: The 4-digit numbers from 1000 through 1999 and from 2000 through 2707 represent the strings BAAA, BAAB, BAAC, …, BJJJ and CAAA, CAAB, CAAC, …, CHAH, respectively, which all come before CHAI alphabetically. That’s a total of 1000 + 708 = 1708 strings.

Altogether, 2 + 18 + 171 + 1708 = 1899 strings appear before CHAI in the list.


Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2