Chapter 2 Numbers and Sequences Ex 2.1
1. Find all positive integers which when divided by 3 leaves remainder 2.
Solution:
By Euclid's division lemma, $a = bq + r$, $0 \le r \lt b$
$a = 3q + 2$, where q=0,1,2...
∴ The positive integers are 2, 5, 8, 11,…
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:
By Euclid's division lemma, $a = bq + r$, $0 \le r \lt b$
$532 = 21 × 25 + 7$
The remainder is 7.
No. of completed rows = 25, left over flower pots = 7 pots.
3. Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let $n,\: n+1$ be two consecutive positive integers.
$n(n+1)=n^2+n$
Case 1:
when n is odd, $n^2$ is also odd.
$\implies n^2+n$ is even.
Case 2:
when n is even, $n^2$ is also even.
$\implies n^2+n$ is even.
Hence the product of two consecutive positive integers is divisible by 2.
4. When the positive integers be 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:
$a = 13 q_1 + 9$
$b = 13 q_2 + 7$
$c = 13 q_3 + 10$
$a + b + c = 13 q_1 + 9 + 13 q_2 + 7 + 13 q_3 + 10$
$a+b+c=13(q_1+q_2+q_3)+26$
$a+b+c=13[q_1+q_2+q_3+2]$
Hence, a + b + c is divisible by 13.
5. Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Solution:
Let the integer be $x$.
Case 1: When $x$ is odd.
$x=2n+1$
$\implies x^2=(2n+1)^2$
$\implies x^2=4n^2+4n+1$
$\implies x^2=4(n^2+n)+1$
Hence, When divided by 4 it leaves a remainder 1.
Case 2: When $x$ is even.
$x=2n$
$\implies x^2=4n^2$
Hence, When divided by 4 it leaves a remainder 0.
Hence it is proved.
6. Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
Solution:
$412 = 340 × 1 + 72$
$340 = 72 × 4 + 52$
$72 = 52 × 1 + 20$
$52 = 20 × 2 + 12$
$20 = 12 × 1 + 8$
$12 = 8 × 1 + 4$
$8 = 4 × 2 + 0$
Therefore H.C.F. of 340 and 412 is 4.
(ii) 867 and 255
Solution:
$867 = 255 × 3 + 102$
$255 = 102 × 2 + 51$
$102 = 51 × 2 + 0$
Therefore the H.C.F. of 867 and 255 is 51.
(iii) 10224 and 9648
Solution:
$10224 = 9648 × 1 + 576$
$9648 = 576 × 16 + 432$
$576 = 432 × 1 + 144$
$432 = 144 × 3 + 0$
Therefore H.C.F. of 10224 and 9648 is 144.
(iv) 84, 90 and 120
Solution:
$90 = 84 × 1 + 6$
$84 = 6 × 14 + 0$
∴ The H.C.F. of 84 and 90 is 6.
$120 = 6 × 20 + 0$
Therefore H.C.F. of 84, 90 and 120 is 6.
7. Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Solution:
$1230-12=1218$
$1926-12=1914$
By Euclid’s division algorithm
$1914 = 1218 × 1 + 696$
$1218 = 696 × 1 + 522$
$696 = 522 × 1 + 174$
$522 = 174 × 3 + 0$
∴ HCF of 1218 and 1914 is 174
8. If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Solution:
$60 = 32 × 1 + 28$
$32 = 28 × 1 + 4$
$28 = 4 × 7 + 0$
∴ H.C.F. of 32 and 60 is 4.
$4=32x+60y$
$⇒64-60=32x+60y$
$⇒(32 \times 2) -(60 \times 1)=32x+60y$
$⇒[32\times 2]+[60 \times (-1)]=32x+60y$
∴ x = 2 and y = -1
9. A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Solution:
Let the positive integer be $x$
$x = 88 × q + 61$
$61 = 11 × 5 + 6$
∴ The remainder is 6.
10. Prove that two consecutive positive integers are always coprime.
Solution:
Let the two consecutive positive integers be $x, \: x+1$, where $x+1 > x$.
By Euclid's division algorithm, $a=bq+r$
$x+1=x \times 1 + 1$
$x=1 \times x + 0$
∴ H.C.F. of the two consecutive numbers is 1.
∴ Two consecutive positive integers are always coprime.