CBSE Class 8 Maths Exemplar Chapter 13: Playing with Numbers - NCERT Solutions

The generalized form of a number expresses it as the sum of the products of its digits with their respective place values. For a four-digit number abdc, the digit 'a' is in the thousands place, 'b' is in the hundreds place, 'd' is in the tens place, and 'c' is in the units place. Therefore, the generalized form is:

$$1000\; \backslash times\; a\; +\; 100\; \backslash times\; b\; +\; 10\; \backslash times\; d\; +\; c$$

This corresponds to option (c).

For a two-digit number xy, the digit 'x' is in the tens place and the digit 'y' is in the units place. To express this in its generalized form, we multiply the tens digit by 10 and add the units digit. Thus, the generalized form is:

$$10\; \backslash times\; x\; +\; y$$

This matches option (b).

The expression $1000a\; +\; 10b\; +\; c$ represents a number where 'a' is the digit in the thousands place, 'b' is the digit in the tens place, and 'c' is the digit in the units place. Since there is no term for the hundreds place (or it is implicitly 0), the number is formed by placing 'a' in the thousands position, '0' in the hundreds position, 'b' in the tens position, and 'c' in the units position. This gives us the number aobc.

The usual form is aobc.

Let the three-digit number be abc. Its value can be written in generalized form as $100a\; +\; 10b\; +\; c$. The number formed by reversing its digits, cba, can be written as $100c\; +\; 10b\; +\; a$.

Now, let's find the difference:

$$abc\; -\; cba\; =\; (100a\; +\; 10b\; +\; c)\; -\; (100c\; +\; 10b\; +\; a)$$

$$=\; 100a\; +\; 10b\; +\; c\; -\; 100c\; -\; 10b\; -\; a$$

$$=\; (100a\; -\; a)\; +\; (10b\; -\; 10b)\; +\; (c\; -\; 100c)$$

$$=\; 99a\; -\; 99c$$

$$=\; 99(a\; -\; c)$$

The expression $abc\; -\; cba$ is equal to $99(a\; -\; c)$. This means the difference is always a multiple of 99. Since 99 is a multiple of 9, 11, and 33 (as $99\; =\; 9\; \backslash times\; 11$, $99\; =\; 3\; \backslash times\; 33$), the expression $abc\; -\; cba$ will always be divisible by 9, 11, and 33.

However, 18 is not a factor of 99. Therefore, $abc\; -\; cba$ is not always divisible by 18.

The number that does not divide abc – cba is 18.

Let the three digits be x, y, and z. The numbers that can be formed by permuting these digits are xyz, xzy, yxz, yzx, zxy, and zyx. However, the question implies the sum of numbers formed by cyclic permutation, which are typically xyz, yzx, and zxy.

Let's consider the generalized form of these three numbers:

$xyz\; =\; 100x\; +\; 10y\; +\; z$

$yzx\; =\; 100y\; +\; 10z\; +\; x$

$zxy\; =\; 100z\; +\; 10x\; +\; y$

Now, let's find the sum of these three numbers:

$$Sum\; =\; (100x\; +\; 10y\; +\; z)\; +\; (100y\; +\; 10z\; +\; x)\; +\; (100z\; +\; 10x\; +\; y)$$

Group the terms with x, y, and z:

$$Sum\; =\; (100x\; +\; x\; +\; 10x)\; +\; (10y\; +\; 100y\; +\; y)\; +\; (z\; +\; 10z\; +\; 100z)$$

$$Sum\; =\; 111x\; +\; 111y\; +\; 111z$$

Factor out 111:

$$Sum\; =\; 111(x\; +\; y\; +\; z)$$

We know that $111\; =\; 3\; \backslash times\; 37$. Therefore, the sum $111(x\; +\; y\; +\; z)$ is always divisible by 111 and its factors.

Let's check the given options:

• 11: 111 is not divisible by 11.
• 33: 111 is not divisible by 33 (since $111\; /\; 33\; \backslash approx\; 3.36$).
• 37: 111 is divisible by 37 ($111\; =\; 3\; \backslash times\; 37$).
• 74: 111 is not divisible by 74.

Thus, the sum of the numbers formed by cyclic permutation of digits x, y, and z is divisible by 37.

The sum is divisible by 37.