SEBA Class 10 Advanced Maths Chapter 3.5 Solutions: Arithmetic of Integers
In this article, we have solved all the questions from Chapter 3.5 of SEBA Class 10 Advanced Mathematics in a simple way. 100% reliable solutions for all the questions of SEBA Class 10 Advanced Maths Chapter 3.5
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SEBA Class 10 Advanced Maths Solutions to Exercise 3.5
Q 1: Determine the integers in between 50 and 100 which are congruent to 1 modulo 4.
Solution:
Let $n$ be the integer which is congruent to $1 \pmod 4$.
Then, $n \equiv 1 \pmod 4$
i.e.,$4 \mid n – 1$
$\implies n – 1 = 4k, \quad k \in \mathbb{Z}$
According to question, $50 < n < 100$
$\implies 50 < 1 + 4k < 100$
$\implies 50 – 1 < 4k < 100 – 1 $
$\implies 49 < 4k < 99$
$\implies 12.25 < k < 24.75$ [Dividing by $4$ ]
From this, we obtain the values of $k$ as:
$k = 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, \text{ and } 24$
Therefore, $n = 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, \text{ and } 97$
Question 2: Find all integers in between 70 and 130 which are congruent to 11 modulo 17.
Solution:
Let $n$ be the integer which is congruent to $11 \pmod{17}$.
Then, $n \equiv 11 \pmod{17}$
i.e., $17 \mid n – 11$
i.e., $n – 11 = 17k, \quad k \in \mathbb{Z}$
i.e., $n = 11 + 17k, \quad k \in \mathbb{Z}$
By the given condition, $70 < n < 130$.
i.e., $70 < 11 + 17k < 130$
i.e., $59 < 17k < 119$
From this, we obtain the values of $k$ as:
Substituting these values of $k$ into $n = 11 + 17k$, we get:
For $k = 4$: $n = 11 + 17(4) = 79$
For $k = 5$: $n = 11 + 17(5) = 96$
For $k = 6$: $n = 11 + 17(6) = 113$
Therefore, $n = 79, 96, \text{ and } 113$.
Question 3: Find all integers congruent to 2 modulo 11.
Solution:
Let $n$ be the integer congruent to $2$ modulo $11$.
i.e., $n \equiv 2 \pmod{11}$
i.e., $11 \mid n – 2$
i.e., $n – 2 = 11k, \quad k \in \mathbb{Z}$
i.e., $n = 2 + 11k, \quad k \in \mathbb{Z}$
Therefore, substituting $k = 0, \pm 1, \pm 2, \dots$, we get:
$n = \dots, -20, -9, 2, 13, 24, \dots$
Solution:
We have, $2^3 = 8 \quad \text{and} \quad 8 \equiv 1 \pmod 7$
$\implies 2^3 \equiv 1 \pmod 7 \quad [\because 7 \mid 8 – 1]$
Now, $(2^3)^{16} \equiv 1^{16} \pmod 7$
$\implies 2^{48} \equiv 1 \pmod 7$
$ \implies 2^{48} \times 2^2 \equiv 1 \times 2^2 \pmod 7$ [multiplying both sides by $2^2$]
$\implies 2^{50} \equiv 4 \pmod 7$
Question 5: Find the remainder when $41^{65}$ is divided by 7.
Solution:
We have, $41 – (-1) = 42$, which is divisible by $7$
$\implies 41 \equiv -1 \pmod 7$
$\implies 41^{65} \equiv (-1)^{65} \pmod 7$
$\implies 41^{65} \equiv -1 \pmod 7$
Since $-1 \equiv 6 \pmod 7 \quad [\because 7 \mid -1 – 6]$
$\implies 41^{65} \equiv 6 \pmod 7$
Thus, the required remainder is $6$.
Question 6: Find the remainder when $100^{100}$ is divided by 101.
Solution:
We have, $100 – (-1) = 101$, which is divisible by $101$
$\implies 100 \equiv -1 \pmod{101}$
$\implies (100)^{100} \equiv (-1)^{100} \pmod{101}$
$\implies 100^{100} \equiv 1 \pmod{101}$
Thus, the required remainder is $1$.
Question 7: Find the remainder when $\lfloor 1 + \lfloor 2 + \lfloor 3 + \dots + \lfloor 50$ is divided by 15.
Solution:
We have, $15 = 5 \times 3$
Now, $\lfloor 5 = 5 \times 4 \times 3 \times 2 \times 1 = 120 = 15 \times 8 $
Therefore, $15 \mid \lfloor 5$
Similarly, $15 \mid \lfloor 6, \ 15 \mid \lfloor 7, \ \dots, \ 15 \mid \lfloor 50$
In other words,
$\lfloor 5 \equiv 0 \pmod{15}$
$\lfloor 6 \equiv 0 \pmod{15}$
$\dots$
$\lfloor 50 \equiv 0 \pmod{15}$
So, $\lfloor 1 + \lfloor 2 + \lfloor 3 + \lfloor 4 + \lfloor 5 + \dots + \lfloor 50$
$\equiv \lfloor 1 + \lfloor 2 + \lfloor 3 + \lfloor 4 + 0 + 0 + \dots + 0 \pmod{15}$
$\equiv 1 + 2 + 6 + 24 \pmod{15}$
$\equiv 33 \pmod{15}$
$\equiv 3 \pmod{15} \quad [\text{since } 33 = 15 \times 2 + 3]$
Thus, the required remainder is $3$.
Question 8: For any odd integer $a$, show that $a^2 \equiv 1 \pmod 8$.
Solution:
By Division Algorithm, any integer $a$ can be expressed as $4k, 4k+1, 4k+2$ or $4k+3$, $k$ being any integer. Since $a$ is odd, $a = 4k+1$ or $4k+3$.
Case I: When $a = 4k + 1$
$\implies a^2 = (4k + 1)^2 = 16k^2 + 8k + 1$
$\implies a^2 = 8(2k^2 + k) + 1$
$\implies a^2 – 1 = 8(2k^2 + k)$
$\implies 8 \mid a^2 – 1$
$\implies a^2 \equiv 1 \pmod 8$
Case II: When $a = 4k + 3$
$\implies a^2 = (4k + 3)^2 = 16k^2 + 24k + 9$
$\implies a^2 = 16k^2 + 24k + 8 + 1$
$\implies a^2 = 8(2k^2 + 3k + 1) + 1$
$\implies a^2 – 1 = 8(2k^2 + 3k + 1)$
$\implies 8 \mid a^2 – 1$
$\implies a^2 \equiv 1 \pmod 8$
Hence proved.
Question 9: If the first January of 1880 was Monday, what was the day on 20 June of that year?
Solution:
Since the year 1880 is a leap year, February has 29 days.
Number of days from 1st January to 20th June, 1880 (excluding the first day):
Jan = $30$, Feb = $29$, Mar = $31$, Apr = $30$, May = $31$, June = $20$
Total days = $30 + 29 + 31 + 30 + 31 + 20 = 171$ days
Now, $171 = 24 \times 7 + 3$
$\implies 171 – 3 = 24 \times 7$
$\implies 7 \mid 171 – 3$
$\implies 171 \equiv 3 \pmod 7$
It means that 24 complete weeks and 3 additional days passed.
Therefore, the day on 20th June = Monday + 3 days = Thursday
Question 10: If the first January of 1936 was Wednesday, what was the day on 31st December of the same year?
Solution:
Since the year 1936 is a leap year, it contains 366 days total.
∴ The total days from 1st January 1936 to 31st December 1936 (excluding the first day) = $366 – 1 = 365$ days
Now, $365 = 52 \times 7 + 1$
$\implies 365 – 1 = 52 \times 7$
$\implies 7 \mid 365 – 1$
$\implies 365 \equiv 1 \pmod 7$
So, 365 days means 52 complete weeks plus 1 extra day.
Therefore, The day on 31st December = Wednesday + 1 day = Thursday
Question 11: For any $a, b \in \mathbb{Z}$ and any fixed integer $n > 1$, show that the relation $a \equiv b \pmod n$ is an equivalence relation in $\mathbb{Z}$.
Solution:
To prove that the relation is an equivalence relation, we must show it is reflexive, symmetric, and transitive.
(i) Reflexivity:
For any integer $a \in \mathbb{Z}$, we have $a – a = 0$, which is divisible by $n$.
$\implies n \mid a – a$
$\implies a \equiv a \pmod n \quad \forall a \in \mathbb{Z}$
Thus, the relation is reflexive.
(ii) Symmetry:
Let $a \equiv b \pmod n$
$\implies n \mid a – b$
$\implies a – b = nq, \quad q \in \mathbb{Z}$
$\implies -(a – b) = n(-q), \quad q \in \mathbb{Z}$
$\implies b – a = n(-q)$
$\implies n \mid b – a$
$\implies b \equiv a \pmod n$
Thus, the relation is symmetric.
(iii) Transitivity:
Let $a \equiv b \pmod n$ and $b \equiv c \pmod n$
$\implies n \mid a – b$ and $n \mid b – c$
$\implies a – b = nk_1$ and $b – c = nk_2, \quad k_1, k_2 \in \mathbb{Z}$
Adding the two equations:
$\implies (a – b) + (b – c) = nk_1 + nk_2$
$\implies a – c = n(k_1 + k_2)$
Since $(k_1 + k_2) \in \mathbb{Z}$, it implies:
$\implies n \mid a – c$
$\implies a \equiv c \pmod n$
Thus, the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation in $\mathbb{Z}$. Hence proved.