Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Solution: Let the two consecutive even positive integers be x and x + 2. x > 5, x + 2 > 5, x + (x + 2) < 23 x > 5, x...

Solution: True. Justification: Let a, a + 1 be two consecutive positive integers. By Euclid’s division lemma, we have a = bq + r, where 0 ≤ r < b For b = 2 , we...

Solution: True. Justification: At least one out of every three consecutive positive integers is divisible by 2. Therefore, The product of three consecutive positive integers is divisible by 2. At least one out of every...

Solution: Since x and y are odd positive integers, we have x = 2m + 1 and y = 2n + 1 => x2 + y2 = (2m + 1)2 + (2n +...

Solution: Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer. By Euclid’s division lemma, we have a = bq + r; 0 ≤...

Correct answer is option (C) 0 ≤ r < b

Let the three consecutive integers bex, x+1and x+2. According to the question, x+x+1+x+2 = 51 => 3x+3=51 => 3x+3-3=51-3 [Subtracting 3 from both sides] => 3x=48 => 3x/3 = 48/3 [Dividing both sides by 3] X=16 Hence, first...

Let, a = 2p+1, p∈N and b = 2q+1 , q∈N\(\cup\){0} \(\because\) a > b ⇒ 2p+1 > 2q+1 ⇒ p > q \(\frac{a+b}{2}\) = \(\frac{2p+1+2q+1}{2}\) = p+q+1 \(\frac{a-b}{2}\) = \(\frac{2p+1-(2q+1)}{2}\) = p-q Case-I: \(\frac{a+b}{2}\) is odd which implies p+q+1 is odd \(\Rightarrow\) p+q is even (\(\because\) odd - 1 = even) \(\Rightarrow\) p+q-2q...

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2. Since both the integers are smaller than 10, x + 2 <...

Let x be the smaller of the two consecutive even positive integers. Then, the other integer is x + 2.  Since both the integers are larger than 5,  x > 5 ..........................