This is the fourth part of the Metrobank-MTAP Math Challenge Questions for Grade 7. In this post, we discuss the solutions to number 21-30. You can also read the solutions to **1-10,** **11-20, **and **21 – 30**.

**31.)** If is a prime number not equal to 2, which of the following can be a prime:

, , .

**Solution**

All prime numbers not equal to 2 are odd. If is odd, then and are both even, while is odd. Therefore, only can be a prime.

**Answer:**

**32.)** What must be added to the product of and to obtain ?

**Solution**

The product of and is .

Now, we subtract the result from .

**Answer:**

**33.)** If the greatest common divisor of two numbers is 12 and their least common multiple is 180, what are the possible pairs of numbers?

**Solution:**

The product of the gcd and lcm of two numbers is equal to the product of the numbers. So, we need to examine the factors of 12(180) = 2160 where the gcd is 12 and lcm is 180. We only have two pairs of such numbers, 12 and 180 and 36 and 60.

**Answer:** {12, 180} and {36, 60}

**34.)** Given a positive integer n, arrange the following from least to greatest:

,

**Solution**

Squaring all the expressions, it is easy to see that

.

The details are left as an exercise.

**Answer:** .

**35.)** What is the largest integer which is always a factor on the sum of three consecutive even integers?

**Solution**

Let , , and be three consecutive even integers. Their sum is . Therefore, 6 is always a factor of their sum.

**Answer:** 6

**36.)** If the sum of the first positive 100 integers is 5050, what is the sum of the next 100 integers?

**Solution:** The next 100 integers are 101, 102, 103, …, 200. That is, we add 100 to each one of the first 100 integers. In effect, we added (100) (100) + 5050 = 15050.

**Answer:** 15050

**37.)** With what polynomial must be divided to get a quotient and a remainder of ?

**Solution**

Consider dividing 12 by 7. We get a quotient of 1 and a remainder of 5. From this, we can form the equation, 12/7 = 1 + 5/7. In general, if we divide and get a quotient and a remainder of , we can have the equation

(1).

In the equation above, we are looking for the value for b, the divisor. Therefore, multiplying the equation in (1) by , we get , giving us . If we let, , and , substituting the values in (*), we get .

Answer:

**38.)** If leaves a remainder of when divided by , what is the remainder when is divided by ?

Solution

The modular inverse of is equal to since . This means that . Multiplying by 8, we have . Therefore, .

Answer: 7

**39.)** Find the solution set: .

**Solution**

, ,

, , .

**Answer:** -8 or 14

40.) Compute:

**Solution**

Answer: