# 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 $n$ is a prime number not equal to 2, which of the following can be a prime:

$n + 3$, $2^n - 1$, $n^2 + 1$.

Solution

All prime numbers not equal to 2 are odd. If $n$ is odd, then $n + 3$ and $n^2 + 1$ are both even, while $2^n - 1$ is odd. Therefore, only $2^n - 1$ can be a prime.

Answer: $2^n - 1$

32.) What must be added to the product of $2x - 1$ and $4x +5$ to obtain $4x^2 + 8x + 1$

Solution

The product of $2x - 1$ and $4x + 5$ is $(2x - 1)(4x + 5) = 8x^2 + 6x - 5$.
Now, we subtract the result from $4x^2 + 8x + 1$.

$4x^2 + 8x + 1 - (8x^2 + 6x - 5) = 4x^2 + 8x + 1 - 8x^2 - 6x + 5 = -4x^2 + 2x + 6$

Answer: $-4x^2 + 2x + 6$

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:

$\sqrt{n} + 1$, $\sqrt{n + 1}, n + 1$

Solution

Squaring all the expressions, it is easy to see that

$n + 1 > \sqrt{n} + 1 > \sqrt{n + 1}$.

The details are left as an exercise.

Answer: $n + 1 > \sqrt{n} + 1 > \sqrt{n + 1}$.

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

Solution

Let $2n$, $2n + 21$, and $2n + 2$ be three consecutive even integers. Their sum is $6n + 6 = 6(n + 1)$. Therefore, 6 is always a factor of their sum.

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.

37.) With what polynomial must$3x^4 - 3x^3 + 5x^2 + 2x$ be divided to get a quotient $3x^2 +2$ and a remainder of $4x - 2$?

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 $a/b$ and get a quotient $q$ and a remainder of $r$, we can have the equation

$a/b = q + r/q$ (1).

In the equation above, we are looking for the value for b, the divisor. Therefore, multiplying the equation in (1) by $q$, we get $a = bq + r$, giving us  $b = (a - r)/q (*)$. If we let$a = 3x^4 - 3x^3 + 5x^2 + 2x$, $q = 3x^2 + 2$, and $r = 4x - 2$, substituting the values in (*), we get $x^2 - x + 1$.

Answer: $x^2 - x + 1$

38.) If $7n$ leaves a remainder of $8$ when divided by $9$, what is the remainder when $5n$ is divided by $9$?

Solution

The modular inverse of $7 \mod 9$ is equal to $1$ since $(7)(4) \equiv 1 (\mod 9)$. This means that $\frac{1}{7} \equiv 4 \mod 9$. Multiplying by 8, we have $\frac{8}{7} \equiv 5 \mod 9$. Therefore, $5n = 25 \equiv 7 \mod 9$.

39.) Find the solution set: $|3 - x| = 11$.

Solution

$3 - x = 11$, $-x = 8$, $x = -8$
$-(3 - x) = 11$, $x - 3 = 11$, $x = 14$.

40.) Compute: $\left ( 1 - \frac{2}{3} \right ) \left ( 1 - \frac{2}{5} \right ) \left ( 1 - \frac{2}{7} \right ) \left ( 1 - \frac{2}{9} \right ) \left ( 1 - \frac{2}{75} \right )$

Solution

$\left ( 1 - \dfrac{2}{3} \right ) \left ( 1 - \dfrac{2}{5} \right ) \left ( 1 - \dfrac{2}{7} \right ) \left ( 1 - \dfrac{2}{9} \right ) \left ( 1 - \dfrac{2}{75} \right )$

$= \left (\dfrac{1}{3} \right ) \left (\dfrac{3}{5} \right ) \left (\dfrac{5}{7} \right )\left (\dfrac{7}{9} \right ) \left (\dfrac{73}{75} \right )$

$= \dfrac{73}{(9)(75)}$

$= \dfrac{73}{675}$

Answer: $\dfrac{73}{675}$

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