find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

The smallest number which when increased by 17 is exactly divisible by both 20 and 4 must be such that the number plus 17 is divisible by the least common multiple (LCM) of 20 and 4. The LCM of 20 and 4 is 20, because 20 is the smallest number divisible by both 20 and 4. Let the smallest number be xxx. Then x+17x+17x+17 is divisible by 20, meaning:
x+17=20kx+17=20kx+17=20k
for some integer kkk. To find the smallest positive xxx, take k=1k=1k=1:
x+17=20×1=20x+17=20\times 1=20x+17=20×1=20
x=20−17=3x=20-17=3x=20−17=3 So, the smallest number is 3\boxed{3}3​. 3 increased by 17 gives 20, which is divisible by both 20 and 4. Thus, the answer is 3.