Prove that if x and y are both odd positive integers, then x^2 + y^2 is even but not divisible by 4.

Solution: Let the two consecutive odd positive integers be x and x + 2. x < 10, x + 2 < 10, x + (x + 2) > 11 x + 2 <...

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: 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

n2 - 1 is divisible by 8, if n is an odd integer 

Let, (n – 1) and n be two consecutive positive integers ∴ Their product = n(n – 1) = n 2 −n We know that any positive integer is of the form 2q...

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 ..........................

(b) By looking at all the options, we observe that option (b) is not necessarily true. We know that, a number is divisible by 10 iff it has 0 at...