Exercise 2.1 - Solutions
Problem 1
Find all positive integers, when divided by 3 leaves remainder 2.
Solution
Let 'a' be any positive integer and 'b' be the divisor.
By Euclid's division lemma, we have:
Here, the divisor $b = 3$ and the remainder $r = 2$.
Substituting these values into the lemma, we get:
Since we are looking for positive integers, the quotient 'q' can be any whole number ($q = 0, 1, 2, 3, \dots$).
- If $q = 0$, then $a = 3(0) + 2 = 2$
- If $q = 1$, then $a = 3(1) + 2 = 5$
- If $q = 2$, then $a = 3(2) + 2 = 8$
- If $q = 3$, then $a = 3(3) + 2 = 11$
- and so on...
Problem 2
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
Solution
Let 'a' be the total number of flower pots and 'b' be the number of flower pots in each row.
Here, $a = 532$ and $b = 21$.
We use Euclid's division lemma to find the quotient 'q' (number of completed rows) and the remainder 'r' (flower pots left over).
Substituting the given values:
$532 = 21 \times q + r$
To find 'q' and 'r', we perform the division:
$532 \div 21$ gives a quotient of 25 and a remainder of 7.
So, we can write:
Here, the quotient $q = 25$ and the remainder $r = 7$.
Problem 3
Prove that the product of two consecutive positive integers is divisible by 2.
Solution
Let the two consecutive positive integers be $n$ and $n+1$. We need to consider two cases for the integer $n$.
Case 1: $n$ is an even number.
If $n$ is even, it can be written as $n = 2k$ for some integer $k \ge 1$.
Their product is $n(n+1) = (2k)(2k+1) = 2(k(2k+1))$.
Since the product is a multiple of 2, it is divisible by 2.
Case 2: $n$ is an odd number.
If $n$ is odd, it can be written as $n = 2k-1$ for some integer $k \ge 1$.
The next consecutive integer is $n+1 = (2k-1)+1 = 2k$.
Their product is $n(n+1) = (2k-1)(2k) = 2(k(2k-1))$.
Again, since the product is a multiple of 2, it is divisible by 2.
Problem 4
When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a+b+c is divisible by 13.
Solution
We are given three positive integers a, b, and c. Using Euclid's division lemma, we can express them in terms of the divisor 13.
For integer 'a', when divided by 13, the remainder is 9. So:
$a = 13q_1 + 9 \quad \dots(1)$For integer 'b', when divided by 13, the remainder is 7. So:
$b = 13q_2 + 7 \quad \dots(2)$For integer 'c', when divided by 13, the remainder is 10. So:
$c = 13q_3 + 10 \quad \dots(3)$Here, $q_1, q_2, q_3$ are some integer quotients.
Now, we need to show that their sum, $a+b+c$, is divisible by 13. Let's add the three equations:
$a+b+c = (13q_1 + 9) + (13q_2 + 7) + (13q_3 + 10)$Group the terms with 13 and the remainder terms:
$a+b+c = (13q_1 + 13q_2 + 13q_3) + (9 + 7 + 10)$Factor out 13 from the first group and sum the remainders:
$a+b+c = 13(q_1 + q_2 + q_3) + 26$We can also factor out 13 from 26, since $26 = 13 \times 2$.
$a+b+c = 13(q_1 + q_2 + q_3) + 13(2)$Now, factor out 13 from the entire expression:
Since $(q_1 + q_2 + q_3 + 2)$ is an integer, the sum $a+b+c$ is a multiple of 13.
Problem 5
Prove that the square of any integer leaves the remainder either 0 or 1 when divided by 4.
Solution
Let 'x' be any integer. Any integer is either even or odd. We will examine the square of x in both cases.
Case 1: x is an even integer.
An even integer can be written in the form $x = 2k$, where k is an integer.
Squaring the integer, we get:
We can write this as:
$x^2 = 4(k^2) + 0$This is of the form $4m + 0$, where $m = k^2$. So, when the square of an even integer is divided by 4, the remainder is 0.
Case 2: x is an odd integer.
An odd integer can be written in the form $x = 2k+1$, where k is an integer.
Squaring the integer, we get:
We can factor out 4 from the first two terms:
$x^2 = 4(k^2 + k) + 1$This is of the form $4n + 1$, where $n = k^2 + k$. So, when the square of an odd integer is divided by 4, the remainder is 1.