CBSE Class 10 Mathematics Real Numbers Notes Set 07

H.C.F. and L.C.M. of Decimals

To find the L.C.M. and H.C.F. of decimal numbers:
1. Convert every decimal to a number with the same number of decimal places.
2. Remove the decimal point and find the H.C.F. or L.C.M. of the resulting whole numbers.
3. In the answer put the decimal point back at the same number of places as in step 1.

Example 1. Find the L.C.M. and H.C.F. of 0.34 and 2.7.
Answer:
0.34 and 2.7 are written as 0.34 and 2.70. Removing the decimal point we get 34 and 270.
H.C.F. (34, 270) = 2 and L.C.M. (34, 270) = 270 × 17 = 4590.
Putting the decimal point back after two places: H.C.F. (0.34, 2.7) = 0.02 and L.C.M. (0.34, 2.7) = 45.9.

Irrational Numbers in Ascending and Descending Order

Irrational numbers with the same index can be compared directly by comparing their radicands. To compare irrational numbers with different indices, convert them into numbers with the same index and then compare.

Writing irrational numbers in ascending and descending order:
1. To write irrational numbers (with the same index) from smaller to greater is known as ascending order.
2. To write irrational numbers (with the same index) from greater to smaller is known as descending order.

Converting irrational numbers with different indices to the same index:
1. Write the indices of the given irrational numbers.
2. Find the least common multiple of the indices.
3. Make the indices the same with the help of the L.C.M. found.
4. Compare the radicands.

Example 1. Write \( \sqrt[3]{24},\ \sqrt[3]{56},\ \sqrt[3]{29} \) in ascending order.
Answer:
The indices are the same (3), so we compare the radicands 24, 29 and 56: 24 < 29 < 56.
\( \sqrt[3]{24} < \sqrt[3]{29} < \sqrt[3]{56} \)

Example 2. Write \( \sqrt{3},\ \sqrt[3]{4},\ \sqrt[4]{2} \) in descending order.
Answer:
The L.C.M. of the indices 2, 3 and 4 is 12.
\( \sqrt{3} = 3^{1/2} = 3^{6/12} = (3^6)^{1/12} = \sqrt[12]{729} \)
\( \sqrt[3]{4} = 4^{1/3} = 4^{4/12} = (4^4)^{1/12} = \sqrt[12]{256} \)
\( \sqrt[4]{2} = 2^{1/4} = 2^{3/12} = (2^3)^{1/12} = \sqrt[12]{8} \)
All the numbers now have the same index, and 729 > 256 > 8. So \( \sqrt{3} > \sqrt[3]{4} > \sqrt[4]{2} \).

Unit Digit of a Number Raised to Any Power

1. Numbers ending in 0, 1, 5 or 6. Any power of a number ending in 0, 1, 5 or 6 ends in the same digit.

Examples: (i) The unit digit of \( 30^{35} \) is 0. (ii) The unit digit of \( 71^{13} \) is 1. (iii) The unit digit of \( 35^{12} \) is 5. (iv) The unit digit of \( 26^{23} \) is 6.

2. Numbers ending in 4 or 9.

Examples: (i) The unit digit of \( 74^{53} \) is 4. (ii) The unit digit of \( 74^{24} \) is 6. (iii) The unit digit of \( 69^{24} \) is 1. (iv) The unit digit of \( 69^{33} \) is 9.

3. Numbers ending in 2, 3, 7 or 8. The unit digits of the powers of 2, 3, 7 and 8 repeat after every cycle of 4 steps. To find the unit digit of any power, divide the power by 4, look at the remainder (0, 1, 2 or 3) and use the table.

Example 1. Find the unit digit of \( 87^{25} \).
Answer:
The remainder when 25 is divided by 4 is 1. In the table, for the unit digit 7 and remainder 1 the unit digit is 7.
So the unit digit of \( 87^{25} \) is 7.

Example 2. Find the unit digit of \( 72^{32} \).
Answer:
The remainder when 32 is divided by 4 is 0. In the table, for the unit digit 2 and remainder 0 the unit digit is 6.
So the unit digit of \( 72^{32} \) is 6.

Last Two Digits of a Product

If \( X \times Y \) is the product where \( X = \ldots ab \) and \( Y = \ldots cd \), the last two digits of the product can be found as follows.
Step 1: The unit digit of \( X \times Y \) is the unit digit of \( b \times d \). The excess digit, if any, is carried to the next step.
Step 2: Find \( a \times d + c \times b \).
Step 3: Add the excess digit of Step 1 to the result of Step 2. The unit digit of this sum is the ten's digit of the product.

Example 1. Find the last two digits of \( 5647 \times 6384 \).
Answer:
Step 1: \( 7 \times 4 = 28 \). The unit digit of the product is 8 and the excess digit is 2.
Step 2: \( a \times d + c \times b = 4 \times 4 + 8 \times 7 = 72 \).
Step 3: 72 + 2 = 74, so the ten's digit is 4.
The last two digits of the product are 48.

Last Two Digits of a Number Raised to Some Power

Let the number be \( X^Y \).

Case 1: the unit digit of X is 1. The unit digit of \( X^Y \) is 1. The ten's digit is the unit digit of the product of the ten's digit of X and the unit digit of Y.

Example 1. Find the last two digits of \( 71^{23} \).
Answer:
The unit digit is 1. The ten's digit is the unit digit of 7 × 3 = 21, that is 1.
The last two digits are 11.

Case 2: the unit digit of X is 3, 7 or 9. First convert X into a number whose unit digit is 1 and then use Case 1.
For 3 and 7: \( X^Y = (X^4)^{Y/4} = (X_1)^{Y/4} \), where \( X_1 = X^4 \).
For 9: \( X^Y = (X^2)^{Y/2} = (X_1)^{Y/2} \), where \( X_1 = X^2 \).

Example 2. Find the last two digits of \( 17^{28} \).
Answer:
The unit digit of 17 is 7, so \( 17^{28} = (17^4)^7 = (83521)^7 \).
The unit digit of \( (83521)^7 \) is 1. The ten's digit is the unit digit of 2 × 7 = 14, that is 4.
The last two digits are 41.

Case 3: the unit digit of X is 2, 4, 6 or 8. Use the following facts.
1. \( 2^{10} \) ends in 24.
2. \( 24^{n} \) ends in 24 when n is an odd positive integer.
3. \( 24^{n} \) ends in 76 when n is an even positive integer.
4. \( 76^{n} \) ends in 76 for every natural number n.

Example 3. Find the last two digits of \( 2^{1024} \).
Answer:
\( 2^{1024} = 2^{1020} \times 2^4 = (2^{10})^{102} \times 2^4 \).
\( 2^{10} \) ends in 24, and a number ending in 24 raised to the even power 102 ends in 76. So \( (2^{10})^{102} \) ends in 76. Also \( 2^4 = 16 \).
The last two digits of 76 × 16 are 16. So the last two digits of \( 2^{1024} \) are 16.

Case 4: the unit digit of X is 5. Use the table. (The last two digits depend on the ten's digit of X and on whether the power Y is odd or even.)

Example 4. Find the last two digits of \( (235)^{237} \).
Answer:
The digit at the ten's place is 3, which is odd, and the power 237 is also odd.
So \( (235)^{237} \) ends in 75.

Identities Related to \( a^n + b^n \) or \( a^n - b^n \)

1. \( a^n - b^n \) is always divisible by \( a - b \).
2. \( a^n - b^n \) is divisible by \( a + b \) if n is even.
3. \( a^n + b^n \) is divisible by \( a + b \) if n is odd.

Example 1. Find the remainder when \( 15^{16} - 6^{16} \) is divided by 21.
Answer:
Here n = 16 is even, so \( 15 + 6 = 21 \) divides \( 15^{16} - 6^{16} \).
The remainder is 0.

Highest Power of a Number in a Factorial

Factorial. The factorial of a non-negative integer n is denoted by n! and is defined as \( n! = n(n-1)(n-2)(n-3)\ldots 3 \cdot 2 \cdot 1 \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \) and \( 2! = 2 \times 1 \). Note that 0! = 1.

Greatest integer function. The greatest integer function of a real number x is denoted by [x] and is defined as the greatest integer less than or equal to x. For example, [3.2] = 3, [3] = 3 and [−1.5] = −2.

Highest power of a prime in a factorial. If p is a prime number, the highest power of p in n! is \( \left[\frac{n}{p}\right] + \left[\frac{n}{p^2}\right] + \left[\frac{n}{p^3}\right] + \left[\frac{n}{p^4}\right] + \cdots \), where [ ] is the greatest integer function.

Highest power of a composite number in a factorial. The highest power of a composite number a in n! is the highest power of the highest prime factor of a in n!, when a is a product of different primes. More generally, the highest power of \( p^a \) (p prime) in n! is \( \left[\dfrac{\text{highest power of } p \text{ in } n!}{a}\right] \).

Example 1. Find (i) the maximum power of 3 in 40!, (ii) the maximum power of 6 in 50!, (iii) the maximum power of 9 in 40!.
Answer:
(i) \( \left[\frac{40}{3}\right] + \left[\frac{40}{9}\right] + \left[\frac{40}{27}\right] = 13 + 4 + 1 = 18 \). The maximum power of 3 in 40! is 18.
(ii) \( 6 = 2 \times 3 \), and 3 is the highest prime factor of 6. Maximum power of 3 in 50! = \( \left[\frac{50}{3}\right] + \left[\frac{50}{9}\right] + \left[\frac{50}{27}\right] = 16 + 5 + 1 = 22 \). The maximum power of 6 in 50! is 22.
(iii) \( 9 = 3^2 \). Maximum power of 3 in 40! is 18, so the maximum power of 9 is \( \left[\frac{18}{2}\right] = 9 \). The maximum power of 9 in 40! is 9.