The 2016 Raytheon MATHCOUNTS National Competition was held May 7-10, 2016 in Washington, DC. Find out more about the MATHCOUNTS Competition Series at www.mathcounts.....
@@someperson5137 same; i could answer some questions faster than these guys, but if i was actually on stage, i would probably just freeze due to the amount of pressure
@@w7shtheworst Tbh, that's not really a good way to choose lol I mean, standardized testing is not the same as math competitions Also, I thought there was a school round? We used those
Andrew Huycke I was supposed to do that this year but my sponsor had no Internet the night she planned to enroll us but I want to do it next year and hopefully move up to international but the Usamo looks difficult
"Four oh three one" is *NOT* the correct answer, because that is not one of the ways a number can be said. Here are a couple of the allowable ways: "four thousand thirty-one" and "forty hundred thirty-one." The second digit from the left is not a letter.
I check each of the questions got no wrong answer so far. but I consumed much more time to understand the questions and set up the math models then calculating, however speed matters. These kid are human computers
The question said that the answer could be expressed in the form of a^b and to find a + b. It did not ask for the number itself which would be huge and unreasonable to compute.
the thing is that luke could easily have won 3 consecutive times, it was only a question short of winning 2016, having won 2017 and 2018, he missed the question when he said 1976 and the answer was 1978.
Easily? Kid, Luke choked at the end of 2016 and couldn’t even answer a simple last question on division and could answer everything else in like a second. It’s called choking in pressure
I teach Indian and Chinese kids for work, Chinese kids are far smarter. Money is another factor, Chinese parents invest on quality math class while Indian parents send their kids to kumon. Indian win spelling bee coz it doesn’t require investment. Chinese kids play instruments far better than Indian too.
20:00 -- 20:50 Both Mathletes missed this question. (a^2 + b^2)^2 = a^4 + 2 * a^2 * b^2 + b^4, or 15^2 = 153 + 2 * a^2 * b^2. Hence, a^2 * b^2 = 36. (Note: It is NOT that a*b = 36.) (a*b)^2 = 36 where a and b are real numbers. So a*b = 6 or -6 and the correct answer is -6. (Note: IF we CHANGE this problem INTO: "........ a + b = 15 and a^2 + b^2 = 153, .........", THEN we WILL get a*b = 36.)
25:00 - 25:48 Both Mathletes missed this question. The area of square ABCD = (AB)^2 = (BC)^2. We are going to compute (AB)^2 or (BC)^2. First, let’s do some analysis: Only two values, 2 and 5, are known, so we must use these two values. Which part is 2? Which part is 5? Well, from A(2,5), draw a line perpendicular to the y-axis, this line intersects the y-axis at a point and let’s call this point E. Then, AE = 2 and OE = 5, where O is the origin. Triangles AEB is a RIGHT triangle; if we also know BE, then we know (AB)^2 and we are done. Well, BE = OE - BO = 5 - BO , so we must find BO. Could BO be 2?? Solution: The two RIGHT triangles AEB and BOC are congruent***. So BO = AE = 2. Then, OC = BE = OE - BO = 3. Thus, in the right triangle AEB, (AB)^2 = 3^2 + 2^2 = 13. (Or, similarly, in the right triangle BOC, (BC)^2 = 3^2 + 2^2 = 13.) [Note that angle ABC is a right angle. *** The two RIGHT triangles AEB and BOC are congruent because (i) both are RIGHT triangles, (ii) hypotenuse AB and hypotenuse BC are equal in length, and (iii) measure of angle ABE = 90 degrees - measure of angle CBO = measure of angle BCO.]
@@nathanshan8026 No, mathcounts copies lots of problems from the AMCs; take for example #21 on national spint 2019, it is a copy of #14 from 2018 AMC 10 with different numbers