Gifted and Talented
It’s hard to believe it’s May. What a great year it has been at Bacon Elementary. In the world of GT we have many exciting things happening this month!
On Wednesday, May 1st we have 20 Bacon “mathletes” participating in the PSD Mighty Math Minds Competition. There will approximately 50 teams at the competition, and we will have four teams representing Bacon. Go Bulldogs!! This is a fun night of problem solving, team work, and pizza (of course)!
Here’s an example of the types of questions the kids will be working to solve. This is from the Math Olympiad Website (https://www.moems.org/ sample_files/SampleE.pdf). Give it a try! In the following cryptarithm, each different letter represents a different digit in the 6-digit numbers. If B ≠ 0, what is the least sum possible?
+ BUTTER *The answer is below…Don’t’ peek!
In addition to Mighty Math Minds, May is also the time of year when our gifted students review their Advanced Learning Plans. Students set both academic and affective (social/emotional goals) at the beginning of the year. I meet with students each trimester to review their goals and they also partake in an affective lesson with our school counselor, Mrs. Eaton. At this month’s meeting students will determine if they achieved their goal for the year. I’m always impressed with students thoughtful and honest reflections of their progress.
The dates/times of our GT Goal Review are as follows:
Tuesday, May 7th
4 th Grade: 9:10 – 10:10
3 rd Grade: 10:10 – 11:10
5 th Grade: 11:10 – 12:10
I have worked with our 5th grade GT students since they were in 3rd grade. I will miss them all so much and I wish them nothing but success in middle school. To all my 5th graders…You have so much to offer the world. Go get ‘em!
ANSWER: Since the question asks for the least sum possible, start with the hundred-thousands place and assign B the least possible value, which is 1, since B ≠ 0. Assign the letters in the ten-thousands place with the least remaining unused digits. Since E repeats in the tens place, assign E = 0. Then assign 2, 3, and 4 to I, A, and U in some order. As a result, the next least available digits for T and R are 5 and 6, respectively. Therefore, the least possible sum is 512,024.