Warning: Serious answer follows... if you were hoping for more "sack" responses, you'll be disappointed.
This is a classic "Chinese Remainder Theorem"...
The problem requires some n where:
n ≡ 3 (mod 17)
n ≡ 10 (mod 16)
n ≡ 0 (mod 15)
If you're not familiar with modulo notation, the first one says that some number n and 3 differ by 17... or in words we're taught in third grade, n divided by 17 has a remainder of 3.
OK... so the approach is to find some x, y, and z such that n = x + y + z with the following constraints:
x ≡ 3 (mod 17), which satisfies the first requirement and y and z are multiples of 17 (so adding them doesn't change the remainder)
y ≡ 10 (mod 16), which satisfies the second requirement and x and z are multiples of 16 (so adding them doesn't change the remainder)
z ≡ 0 (mod 15), which satisfies the third requirement and x and y are multiples of 15 (so adding them doesn't change the remainder)
And so this really means:
x ≡ 3 (mod 17) and x is a multiple of (15)(16) = 240
y ≡ 10 (mod 16) and y is a multiple of (15)(17) = 255
z ≡ 0 (mod 15)... and this is already satisfied by x and y being a multiple of 15, so we can forget about it!
OK then... let's find an x that works:
240 / 17 = 14 R2
480 / 17 = 28 R4
720 / 17 = 42 R6
960 / 17 = 56 R8
1200 / 17 = 70 R10
1440 / 17 = 84 R12
1680 / 17 = 98 R14
1920 / 17 = 112 R16
2160 / 17 = 127 R1 (126 R18)
2400 / 17 = 141 R3 (140 R20) <---- this is the one!
And then let's find a y that works:
255 / 16 = 51 R15
510 / 16 = 31 R14
765 / 16 = 47 R13
1020 / 16 = 63 R12
1275 / 16 = 79 R11
1530 / 16 = 95 R10 <---- this is the one!
So... n = 2400 + 1530 = 3930