TMUA Number Theory 4

Question 1

For which value of k is 3k + 2 divisible by 7?

Accepted Answer

k = 8

Answer

3(8)+2=26 is not divisible by 7; testing k=5: 17 ✗; k=8: 26 ✗; k=11: 35 ✓; so k=11.

Question 2

What is the largest prime factor of 2310?

Accepted Answer

11

Answer

2310 = 2 × 3 × 5 × 7 × 11; the largest prime factor is 11.

Question 3

If n = 2^a × 3^b has exactly 12 divisors, which of the following is a possible (a, b) pair?

Accepted Answer

(3, 2)

Answer

(a+1)(b+1)=12; for (3,2): 4×3=12 ✓; for (5,1): 6×2=12 ✓ — but (3,2) is listed first and valid, so it is a correct answer.

Question 4

What is the remainder when 1! + 2! + 3! + … + 100! is divided by 12?

Accepted Answer

9

Answer

For n ≥ 4, n! is divisible by 12; so the sum mod 12 = (1+2+6+24) mod 12 = 33 mod 12 = 9.

Question 5

How many pairs of positive integers (m, n) with m < n satisfy gcd(m, n) = 5 and m + n = 50?

Accepted Answer

2

Answer

Write m=5a, n=5b with gcd(a,b)=1 and a+b=10, a<b; coprime pairs: (1,9),(3,7) — that's 2 pairs.

(TMUA) Test of Mathematics for University Admission Practice Test