### Instruction: Suppose there is a certain number that leaves a remainder of 2 when divided by 5 and also leaves a remainder of 3 when divided by 7. What is the smallest positive integer that satisfies this condition?### Response: Let's call the smallest positive integer that satisfies this condition x. We can write the given information as:

x ≡ 2 (mod 5)
x ≡ 3 (mod 7)

We can use the Chinese Remainder Theorem to solve this system of congruences. First, we find the modular inverses:

Let N1 = 7 and N2 = 5, then N = N1 * N2 = 35.

n1 = N/N1 = 35/7 = 5
n2 = N/N2 = 35/5 = 7

Now we need to find the modular inverses:

n1 * a1 ≡ 1 (mod N2)
5 * a1 ≡ 1 (mod 5)

Since 5 is already 0 (mod 5), we can ignore this equation as it will not affect the final result.

n2 * a2 ≡ 1 (mod N1)
7 * a2 ≡ 1 (mod 7)

Trying values for a2, we find that a2 = 1 satisfies the equation:

7 * 1 ≡ 1 (mod 7)

Now we can find x:

x = (2 * n1 * a1 + 3 * n2 * a2) % N
x = (2 * 5 * 0 + 3 * 7 * 1) % 35
x = (0 + 21) % 35
x = 21

So, the smallest positive integer that satisfies the given conditions is x = 21.