CBSE Class 10 Mathematics Chapter 1: Real Numbers NCERT Solutions

We use Euclid's division algorithm to find the HCF for each pair of numbers.

1. 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 divisor (90) and the remainder (45):

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

   The remainder is now zero. The divisor at this stage is 45. Therefore, the HCF of 135 and 225 is 45.

2. 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 divisor at this stage is 196. Therefore, the HCF of 196 and 38220 is 196.

3. 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 divisor (102) and the remainder (51):

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

   The remainder is now zero. The divisor at this stage is 51. Therefore, the HCF of 867 and 255 is 51.

Let 'a' be any positive odd integer. According to Euclid's division lemma, for any integer 'a' and a positive integer 'b', there exist unique integers 'q' (quotient) and 'r' (remainder) such that $a\; =\; bq\; +\; r$, where .

Let's choose the divisor $b\; =\; 6$. Then, any positive integer 'a' can be expressed in one of the following forms:

$a\; =\; 6q\; +\; 0$

$a\; =\; 6q\; +\; 1$

$a\; =\; 6q\; +\; 2$

$a\; =\; 6q\; +\; 3$

$a\; =\; 6q\; +\; 4$

$a\; =\; 6q\; +\; 5$

where 'q' is some integer and the remainder 'r' can be 0, 1, 2, 3, 4, or 5.

Now, we need to show that only the forms $6q\; +\; 1$, $6q\; +\; 3$, and $6q\; +\; 5$ represent odd integers. An integer is odd if it is not divisible by 2, meaning it can be written in the form $2k\; +\; 1$ for some integer k.

• Consider $6q$: This can be written as $2(3q)$. Since $3q$ is an integer, $6q$ is an even number.
• Consider $6q\; +\; 1$: This can be written as $2(3q)\; +\; 1$. Let $k\_1\; =\; 3q$. Then $6q\; +\; 1\; =\; 2k\_1\; +\; 1$, which is an odd number.
• Consider $6q\; +\; 2$: This can be written as $2(3q\; +\; 1)$. Since $3q\; +\; 1$ is an integer, $6q\; +\; 2$ is an even number.
• Consider $6q\; +\; 3$: This can be written as $(6q\; +\; 2)\; +\; 1\; =\; 2(3q\; +\; 1)\; +\; 1$. Let $k\_2\; =\; 3q\; +\; 1$. Then $6q\; +\; 3\; =\; 2k\_2\; +\; 1$, which is an odd number.
• Consider $6q\; +\; 4$: This can be written as $2(3q\; +\; 2)$. Since $3q\; +\; 2$ is an integer, $6q\; +\; 4$ is an even number.
• Consider $6q\; +\; 5$: This can be written as $(6q\; +\; 4)\; +\; 1\; =\; 2(3q\; +\; 2)\; +\; 1$. Let $k\_3\; =\; 3q\; +\; 2$. Then $6q\; +\; 5\; =\; 2k\_3\; +\; 1$, which is an odd number.

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