practice problems pending

Q1. How many two digit numbers have exactly 4 positive factors? (Here 1 and the number (n) are also considered as factors of (n).)

S1.
Only 2 possibilities:
1. p^3 where p is prime
2. p*q where p,q are prime.

p^3 only one 3^3: 1
p*q we can construct
2*(5 to 47 primes): 13
3*(5 to 31): 9
5*(7 to 19): 5
7*(11 to 13): 2

Total: 30 = answer.

Q2. Find the number of pairs ((a,b)) of natural numbers such that (b) is a 3-digit number, (a+1) divides (b-1), and (b) divides (a^2+a+2).
S2.
b-1 = k(a+1)
b = k(a+1) + 1
k(a+1) + 1 divides a^2 + a +2
=> ak + k + 1 divides k(a^2 + a +2) = 2k + ak + a^2.k = 2k + a(k + ak + 1 - 1) = 2k - a + a(k + ak + 1)
=> ak + k + 1 divides 2k - a
Now 3 cases from here:
2k - a = 0 => a = 2k
2k - a > 0 => ak + k + 1 < 2k - a => a(k+1) -k + 1 < 0 => a(k+1) <= k-1 => a <= (k-1)/(k+1) which is not possible since a > 1 and RHS is < 1.
2k - a < 0
=>  ak + k + 1 < a - 2k => a(k-1) + 3k + 1 < 0 which is not possible since LHS > 0

So only possibility is a = 2k => k is positive.
=> b = k(2k+1) + 1 = 2k^2 + k + 1
Does it divide a^2 + a + 2 = 4k^2 + 2k + 2 = 2b. Yes.
So all positive values of k are valid with the constraint that 100 <= b <= 999

=> 100 <= 2k^2 + k + 1 <= 999
=> k ranges from 7 to 22
=> 16 such pairs for (a,b)

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