CBSE Class 10 Mathematics Chapter 1: Real Numbers NCERT Solutions

1. To find the HCF of 135 and 225:

   Since 225 is greater than 135, we apply the division lemma to 225 and 135:

   $$225\; =\; 135\; \backslash times\; 1\; +\; 90$$

   The remainder is 90, which is not zero. So, we apply the division lemma to the divisor (135) and the remainder (90):

   $$135\; =\; 90\; \backslash times\; 1\; +\; 45$$

   The remainder is 45, which is not zero. So, we apply the division lemma to the new divisor (90) and the new remainder (45):

   $$90\; =\; 45\; \backslash times\; 2\; +\; 0$$

   The remainder is now zero. The process stops. The last non-zero remainder, which is the divisor at this stage, is 45.

   Therefore, the HCF of 135 and 225 is 45.

2. To find the HCF of 196 and 38220:

   Since 38220 is greater than 196, we apply the division lemma to 38220 and 196:

   $$38220\; =\; 196\; \backslash times\; 195\; +\; 0$$

   The remainder is zero in the first step. The process stops. The divisor at this stage is 196.

   Therefore, the HCF of 196 and 38220 is 196.

3. To find the HCF of 867 and 255:

   Since 867 is greater than 255, we apply the division lemma to 867 and 255:

   $$867\; =\; 255\; \backslash times\; 3\; +\; 102$$

   The remainder is 102, which is not zero. So, we apply the division lemma to the divisor (255) and the remainder (102):

   $$255\; =\; 102\; \backslash times\; 2\; +\; 51$$

   The remainder is 51, which is not zero. So, we apply the division lemma to the new divisor (102) and the new remainder (51):

   $$102\; =\; 51\; \backslash times\; 2\; +\; 0$$

   The remainder is now zero. The process stops. The last non-zero remainder, which is the divisor at this stage, is 51.

   Therefore, the HCF of 867 and 255 is 51.

Let $a$ be any positive integer. According to Euclid's division algorithm, when $a$ is divided by $b$, we have $a\; =\; bq\; +\; r$, where $q$ is the quotient and $r$ is the remainder such that .

In this case, we consider the divisor $b\; =\; 6$. So, any positive integer $a$ can be expressed in one of the following forms:

$a\; =\; 6q\; +\; 0$, $a\; =\; 6q\; +\; 1$, $a\; =\; 6q\; +\; 2$, $a\; =\; 6q\; +\; 3$, $a\; =\; 6q\; +\; 4$, or $a\; =\; 6q\; +\; 5$, where $q\; \backslash ge\; 0$ is an integer.

Now, let's examine each form to determine if it represents an odd integer:

• $6q$ is an even number because it is divisible by 2 ($6q\; =\; 2\; \backslash times\; (3q)$).
• $6q\; +\; 1$ is an odd number because it can be written as $2(3q)\; +\; 1$, which is in the form $2k\; +\; 1$ where $k\; =\; 3q$.
• $6q\; +\; 2$ is an even number because it is divisible by 2 ($6q\; +\; 2\; =\; 2\; \backslash times\; (3q\; +\; 1)$).
• $6q\; +\; 3$ is an odd number because it can be written as $(6q\; +\; 2)\; +\; 1\; =\; 2(3q\; +\; 1)\; +\; 1$, which is in the form $2k\; +\; 1$ where $k\; =\; 3q\; +\; 1$.
• $6q\; +\; 4$ is an even number because it is divisible by 2 ($6q\; +\; 4\; =\; 2\; \backslash times\; (3q\; +\; 2)$).
• $6q\; +\; 5$ is an odd number because it can be written as $(6q\; +\; 4)\; +\; 1\; =\; 2(3q\; +\; 2)\; +\; 1$, which is in the form $2k\; +\; 1$ where $k\; =\; 3q\; +\; 2$.

Therefore, any positive odd integer must be of the form $6q\; +\; 1$, $6q\; +\; 3$, or $6q\; +\; 5$.