Budding Mathematicians Participate in State Wide MATHCOUNTS Competition

Team Cuffee Zoe C., Joseph M., Luke T. and Adan B. join with Daniel Q., Madel C., Milly A. and Roma T. at the state MATHCOUNTS competition.

Team Cuffee Zoe C., Joseph M., Luke T. and Adan B. join with Daniel Q., Madel C., Milly A. and Roma T.
at the state MATHCOUNTS competition.

On Saturday, March 4, eight Paul Cuffee middle school students participated in the annual Rhode Island MATHCOUNTS competition at the Community College of Rhode Island in Warwick. MATHCOUNTS is a national middle school math competition with a total of about 140,000 students from all 50 states taking part each year. The competition consists of three rounds: two individual rounds and one team round. These problems are difficult: as a point of reference, just 9 of the 189 participants at the Rhode Island competition last year got more than 40% of the first round’s problems correct.

Most of the students on Paul Cuffee’s team have been working on these types of problems in their Math Extensions class throughout this year. In addition, the team members had practices after school in the months preceding the competition. Paul Cuffee’s official team, made up of Adan Berreondo, Zoe Cute, Joseph Malik, and Luke Taylor finished sixth in the state, finishing behind only two public schools (Barrington and Archie Cole, from East Greenwich). They were led by Joseph, who finished 15th in the state. Zoe and Luke also did quite well, with 33rd and 36th best scores, respectively. Four other students competed admirably as individuals: eighth grader Roma Taitwood and seventh graders Milly Asherov, Madel Cabreja, and Daniel Quiroa. The students worked hard to challenge themselves and enjoyed both the preparation and competition. The seventh graders are looking forward to doing even better next year.

Test your own math acuity:
Here is a problem from last year’s Sprint Round: this is one problem of 30 that the students have 40 minutes to complete.

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?

View Answer & Rationale

The answer is 721

Why?

The place to start is the ones digit. The ones digit of the product (91,567) is a 7. When you multiply two numbers, regardless of how many digits the numbers are, the ones digit of the product will be the product of the ones digits of the two numbers. That is, if we multiply xxx,xx3 by xx,xx5, though we obviously don’t know the product, we know the product will have a ones digit of 5 (3×5=15). In this case, with a ones digit of 7, the ones digits of our two numbers have to be either 1 and 7 or 3 and 9–these are the only combinations that give you a 1 in the ones digit. And since the the number is a three digit number that is reversed, our numbers are either 1_7 times 7_1 or 3_9 times 9_3. Doing some quick guess and check, we find out that 721 x 127 = 91,567. The question said that the new number was greater than the original number, so the new number is 721.

by Ben Hall, Title 1 Math, Middle School

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