divisibility rules - 9 is sum of digits! just like 3

p16 of https://drive.google.com/file/d/1rsVo824QIRrGeTfRcXH2AimsOkfi86N-/view?usp=sharing

Factors of Factorials

• ● Mathcounts sets questions involving looking for factors in factorials. The book does factors of 3 in 9! and 99!. Let us look at factors of 10 in 125!
  • ○ Factors of 10 are equivalent to number of zeros to left of decimal point (at end of number)
  • ○ Countdown 2014 asked “13. How many zeros are after the last nonzero digit of 125! ?”
• ● 10 = 2.5 and there are lots of 2’s. We just need to count the number of 5’s in 125!
  • ○ 5 10 15 .. 120 125 gives 25 multiples of 5.
  • ○ But 25 50 75 100 125 all are divisible by 25 and give 2 5’s 5 times
  • ○ 125 gives 3 5‘s once

● This gives 31 factors of 5 and so 31 factors of 10

Key property of unit digits

• ● If you sum two numbers n₁ and n₂ together, the unit digit of the sum is the unit digit of the sum of units digits of n₁ and unit digits of n₂
• ● If you multiply two numbers n₁ and n₂ together, the unit digit of the product is the unit digit of the product of units digits of n₁ and unit digits of n₂
• ● What is 6437 . 7654 ?
• A) 49268796 B) 49268797 or C) 49268798
• ● The answer is C as 7.4 = 28

Perfect Squares and unit digits

O² = 0 ; 1² = 1 ; 2² = 4 ; 3² = 9 ; 4² = 16;

5² = 25; 6² = 36; 7² = 49; 8² = 64; 9² = 81

So perfect squares must end in 0, 1, 4, 5, 6, or 9

The units digits of a perfect square cannot be 2, 3, 7, or 8

p25 about unit digits

part 6 factorial

part 7 modular arithmetic

page 39 geometry! lines, triangles, circle, polygons

Similarity

● The exact similarity of shapes is defined through having same values for angles or equal ratios for corresponding sides

● If last (linear) ratio is L

• ● Ratio of Areas of similar shapes is L²
• ● Ratio of volumes of similar shapes is L³

Mersenne Primes and Perfection

• ● Note a Mersenne prime is a prime number that is one less than a power of two.
• ● If 2^k-1 is a prime number (a Mersenne prime), then MP = 2^k-1(2^k-1) is a perfect number and every even perfect number has this form.
• ● Sum of factors of MP is
  • ○ SF(MP) = (1 + 2 + 2² + .. + 2^k-1) (1 + (2^k-1)) if (2^k-1) is prime
  • ○ The first sum is just (2^k-1) and so SF(MP) = (2^k-1)2^k = 2 MP is perfect
• ● Suppose n = 2^k-1(2^k-1) where(2^k-1) is NOT prime but say (2^k-1) = p₁p₂ where and p₁and p₂ are primes. _{2 k-1}
  • ○ Then SF(n) = (1 + 2 + 2 + .. + 2 )(1 + p₁+ p₂ + p₁p₂ )
  • ○ = (2^k-1)(2^k+ p₁+ p₂ ) which is > 2n proving we need Mersenne primes