1. The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer
No. According to Euclid’s division lemma,
a = 3q + r, where 0 ≤ r < 3 and r is an integer. Therefore, the values of r can be 0, 1 or 2.
2. Can the number 6n, n being a natural number, end with the digit 5? Give reasons.
: No, because 6n = (2 × 3)n = 2n × 3n, so the only primes in the factorization of 6n are 2 and 3, and not 5. Hence, it cannot end with the digit 5.
3. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.
No, because an integer can be written in the form 4q, 4q+1, 4q+2, 4q+3.
4. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.
True, because n (n+1) will always be even, as one out of n or (n+1) must be even
5. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false”? Justify your answer.
True, because n (n+1) (n+2) will always be divisible by 6, as at least one of the factors will be divisible by 2 and at least one of the factors will be divisible by 3.
6. Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.
No. Since any positive integer can be written as 3q, 3q+1, 3q+2,
therefore, square will be 9q2 = 3m, 9q2 + 6q + 1 = 3 (3q2 + 2q) + 1 = 3m + 1, 9q2 + 12q + 3 + 1 = 3m + 1.
7. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.
No. (3q + 1)2 = 9q2 + 6q + 1 = 3 (3q2 + 2q) = 3m + 1.
8. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.
HCF = 75, as HCF is the highest common factor
9. Explain why 3 × 5 × 7 + 7 is a composite number.
3×5×7+7 = 7 (3×5 + 1) = 7 (16), which has more than two factors
10. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
No, because HCF (18) does not divide LCM (380).
11. Without actually performing the long division, find if 987/10500 will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.
Terminating decimal expansion, because 987/ 10500 = 47/ 500 and 500 =53 22
12. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form p/q ? Give reasons.
Since 327.7081 is a terminating decimal number, so q must be of the form 2m.5n;m, n are natural numbers
1. Show that the square of an odd positive integer is of the form 8m + 1, for some whole number m.
Any positive odd integer is of the form 2q + 1, where q is a whole number. Therefore, (2q + 1)2 = 4q2 + 4q + 1 = 4q (q + 1) + 1, (1) q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number.
Therefore, (2q + 1)2 = 4.2 m + 1 = 8 m + 1. [From (1)]
2. Prove that √2 + √3 is irrational.
Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a , where a is rational.
Therefore, √2 = a − √3
Squaring on both sides, we get
2 = a2 + 3 – 2a√ 3
Therefore,
√3= [a2 + 1]/a
Irrational = Rational
Which is a contradiction as the right hand side is a rational number while √3 is irrational.
Hence, √2 + √3 is irrational.
3. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5
Thus we have:
(6 m +1)2 = 36 m2 + 12 m + 1 = 6 (6 m2 + 2 m) + 1 = 6 q + 1, q is an integer
(6 m + 3)2 = 36 m2 + 36 m + 9 = 6 (6 m2 + 6 m + 1) + 3 = 6 q + 3, q is an integer
(6 m + 5)2 = 36 m2 + 60 m + 25 = 6 (6 m2 + 10 m + 4) + 1 = 6 q + 1, q is an integer.
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.
Now Solve these problems
4. Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
5. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
6. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
7. Show that the square of any odd integer is of the form 4q + 1, for some integer q.
8. If n is an odd integer, then show that n2 – 1 is divisible by 8.
9. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.
10. Use Euclid’s division algorithm to find the HCF of 441, 567, 693.
11. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
12. Prove that √3 + √5 is irrational.
13. Show that 12n cannot end with the digit 0 or 5 for any natural number n.
14. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
15. Prove that √ p + √q is irrational, where p, q are primes.
16. Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.
17. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer
18. Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
19. Prove that one of any three consecutive positive integers must be divisible by 3.
Let n,n+1,n+2 be three consecutive positive integers
where n can take the form 3q, 3q+1 or 3q+2.
where n can take the form 3q, 3q+1 or 3q+2.
Case I when n=3q
Then n is divisible by 3
but neither n+1 nor n+2 is divisible by 3.
Then n is divisible by 3
but neither n+1 nor n+2 is divisible by 3.
Case II when n=3q+1
Then n is not divisible by 3.
Then n is not divisible by 3.
n+1 = 3q+1+1 = 3q+2,
which is not divisible by 3.
which is not divisible by 3.
n+2= 3q+1+2 = 3q+3=3(q+1),
which is divisible by3.
which is divisible by3.
Case III when n=3q+2
Then n is not divisible by 3.
Then n is not divisible by 3.
n+1 = 3q+2+1 =3q+3=3(q+1),
which is divisible by 3.
which is divisible by 3.
n+2 = 3q+2+2 = 3q+4,
which is not divisible by 3.
which is not divisible by 3.
Hence, one of n, n+1 and n+2 is divisible by 3.
20. For any positive integer n, prove that n3 – n is divisible by 6.