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.