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:

#38 is hard to explain although I found the answer. If (7n – 8)|9 , then (5n – ?)|9—-

Sol. if n = 5, the hypothesis is correct giving the remainder(?) be 7

Actually, this can be done easily using modulo division. I just want to have a solution accessible to high school students. As of now, I have no time to go back to it yet. Still typing questions from 2015. I will complete the solutions once I have uploaded everything. I’ll try to look at your solution later. Thank you.

This has a lot of grammatical mistakes (or maybe the test paper has. I don’t know, it’s not the same) so I was confused. All in all, this was very very helpful and thank you for providing this info. 🙂

I don’ understnad number 40 why is the answer 73/75…

This is just a simple typo error, it’s 73/675. I fixed it already.

try to fix no.35

no. 35 is asking for 3 consecutive even integers…

Oh, thank you. I missed that. Changed it already.

i think the solution in no. 35 is correct

no, no.35 is asking for the largest integer in the factor

Fixed. Thank you.

waiting for the solution of 38

Done it already using modulo division. Sorry, I was looking for an easier solution, but I can’t find one.

Hello. I’m a student from A science school. I’m sorry to say but the answer in no.35 is 6 coz it says that “three consecutive even integers”, not just integers. Try to solve it again.

Thank you very much for pointing that out. I missed the word “even.” I already changed the solution. 🙂