Tiêu đề 01 Miscellaneous Math Solutions.pdf ✅

01 MISCELLANEOUS MATH SOLUTIONS ▣ 1. Take note that his can be solved using the shift-solve function in a calculator. The solution to be shown is the manual solution. A − 6 8 = 0.001 A − 3 4 = 0.001 (A − 3 4) − 4 3 = 0.001− 4 3 A = 0.001− 4 3 A = 10000

▣ 3. • A. By pattern recognition on the first few powers of 3 and when divided by 5. 3 1 = 3 → Remainder 3 3 2 = 9 → Remainder 4 3 3 = 27 → Remainder 2 3 4 = 81 → Remainder 1 And it repeats for every 4th power. To solve for the remainder when 3 983 is divided by 5, Find the remainder when 983 is divided by 4. 983 ÷ 4 → Remainder 3 Therefore, the remainder when 3 983 is divided by 5 is the same as the remainder when 3 3 is divided by 5. Therefore, the answer is 2 . • B. By Fermat’s Little Theorem. Note that for a prime number p, a positive integer a relatively prime to p satisfies a p−1 ≡ 1 (mod p) Which means that 3 5−1 = 3 4 ≡ 1(mod 5). 3 983 = (3 4 ) 245 ⋅ 3 3 ≡ 1 245 ⋅ 27 (mod 5) = 27 (mod 5) ≡ 2 Note: a ≡ b (mod c) means, b is the remainder when a is divided by c.
▣ 4. • A. By infinite geometric progression. Recall: S∞ = a 1−r where a is the first term and r is the common ratio. 0.38444 ... = 0.38 + 0.00444 ... = 0.38 + 0⏟. 004 + 0 .0004 + 0 .00004 + 0 .000004 + ⋯ a=0.004,r=0.1 = 0.38 + 0.004 1 − 0.1 = 173 450 • B. By algebra. x = 0.384444.. 100x = 38.4444 ... Subtracting the two equations, 100x − x = 38.444 ... − 0.38444 ... 99x = 38.06 x = 38.06 99 x = 173 450