#amc10

[thinkmatrix]

［Math Knowledge Tip 2］ Best math plans for you——https://www.thethinkacademy.com/ThinkMatrix👈👈 Today we are going to learn about Prime Numbers and Composite Numbers✅✅ ⛵️Prime number：A number has only 1 and its own two factors, such a number is called a prime number. The smallest prime number is 2. ⛵️Composite Number：A number that has other factors besides 1 and itself is called a composite number. The smallest composite is 4. 🥰🥰Here are some tips for you: 1.0 and 1are not prime numbers, nor are they composite numbers. 2.The prime numbers within 100🍭🍭 #AMC8 #amc10 #math #AmericanMathmatics #AMC12 #maths #mathslearning #studying #amcexam #amc8exam #AmericanMathmaticsCompetition #studyingpartner #lovemathswithsip https://www.instagram.com/p/CmpGlCsyBpE/?igshid=NGJjMDIxMWI=

[thinkmatrix]

［Math Knowledge Tip 2］ Best math plans for you——https://www.thethinkacademy.com/ThinkMatrix👈👈 Today we are going to learn about Prime Numbers and Composite Numbers✅✅ ⛵️Prime number：A number has only 1 and its own two factors, such a number is called a prime number. The smallest prime number is 2. ⛵️Composite Number：A number that has other factors besides 1 and itself is called a composite number. The smallest composite is 4. 🥰🥰Here are some tips for you: 1.0 and 1are not prime numbers, nor are they composite numbers. 2.The prime numbers within 100🍭🍭 #AMC8 #amc10 #math #AmericanMathmatics #AMC12 #maths #mathslearning #studying #amcexam #amc8exam #AmericanMathmaticsCompetition #studyingpartner #lovemaths https://www.instagram.com/p/CmpGewayfI2/?igshid=NGJjMDIxMWI=

[mymathblog]

AMC 10-2007 (B25)

How many pairs of positive integers (a, b) are there such that a and b have no common factors greater than 1 and a/b + 14b/a is an integer?

(A) 4    (B) 6    (C) 9    (D) 12    (E) infinitely many

We can rewrite the fractions as (9a^2 + 14b^2) / 9ab. Therefore, 3|b. Since a and b are coprime, a|9a^2 + 14b^2, so a|14. Similarly, b|9. However, b cannot be 9 as 81a|9a^2 + 81*14 only has solutions when 3|a, which can’t happen when a|14. So, b=3, and a can be {1, 2, 7, 14}. Checking shows us that all 4 work, so the answer is A (4).

[mymathblog]

AMC 10-2019 (A4)

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

(A) 75    (B) 76    (C) 79    (D) 84    (E) 91

The maximum number of balls we can draw without drawing 15 of the same color comes from this scenario: we draw 14 red, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. This gives us 75 balls. Drawing one more ball will guarantee us 15 of the same color, so the answer is B (76).

[notburch-blog]

Less than a month until the first AMC test and I still know literally nothing at all

[subtlypeachy]

yayy~

apparently this year's top 1% distinguished honor roll was 117 and higher for AMC 10A. I got 117. WHEEEEEEEEEEEEEE 

and for AMC 10B I got 121.5 

I am satisfied with myself

sorry for this bragging post I'm just really happy right now