# Grade 7 MTAP 2015 Questions with Solutions Part 4

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

it is not asking for 3 consecutive even integers

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. 🙂

how can I download a copy of the solutions?