AOB can be 180 at most and 0 at least. 180-0 = 180
2. 203010 [@StupidPersonRomania got correct]
10ac = 1000a+10c+1000
ac = 100a+c+100
a(c-100)-(c-100)=200 [SFFT]
(a-1)(c-100) = 200
(a,c) = (201, 101)
10ac = 203010
3. 103 [@StupidPersonRomania got correct]
(6!+1)/(3!+1)=103. We claim this is smallest.
Suppose that it is not the smallest, and for some a=x and b=y, (x!+1)/(y!+1) < 103 and is an integer. Then 103 > (x!+1)/(y!+1) = 1+ (x!-y!)/(y!+1)=1+y!*((y+1)*y*...*(x-1))/(y!+1) >= 1+y!, so 102>y!, and y = 1, 2, 3, or 4.
y=1: (x!+1)/2 is integer, so x! is odd, so x = 1, but 1 is not larger than 1.
y=2: (x!+1)/3 is integer, but x>2 so x!+1 is not multiple of 3.
y=3: (x!+1)/7 is integer, and x>3. x=4 does not work, x=5 does not work, x=6 was our original solution.
y=4: (x!+1)/25 is integer, and x>4. x=5 does not work, x=6 does not work, but x=7 is too large.
Therefore, there is no solution less than 103, proving the claim.
1. 180 [@StupidPersonRomania got correct]
AOB can be 180 at most and 0 at least. 180-0 = 180
2. 203010 [@StupidPersonRomania got correct]
10ac = 1000a+10c+1000
ac = 100a+c+100
a(c-100)-(c-100)=200 [SFFT]
(a-1)(c-100) = 200
(a,c) = (201, 101)
10ac = 203010
3. 103 [@StupidPersonRomania got correct]
(6!+1)/(3!+1)=103. We claim this is smallest.
Suppose that it is not the smallest, and for some a=x and b=y, (x!+1)/(y!+1) < 103 and is an integer. Then 103 > (x!+1)/(y!+1) = 1+ (x!-y!)/(y!+1)=1+y!*((y+1)*y*...*(x-1))/(y!+1) >= 1+y!, so 102>y!, and y = 1, 2, 3, or 4.
y=1: (x!+1)/2 is integer, so x! is odd, so x = 1, but 1 is not larger than 1.
y=2: (x!+1)/3 is integer, but x>2 so x!+1 is not multiple of 3.
y=3: (x!+1)/7 is integer, and x>3. x=4 does not work, x=5 does not work, x=6 was our original solution.
y=4: (x!+1)/25 is integer, and x>4. x=5 does not work, x=6 does not work, but x=7 is too large.
Therefore, there is no solution less than 103, proving the claim.