CBSE Class 10 Mathematics Real Numbers Notes Set 16

Kinds of Numbers

Natural numbers are the counting numbers 1, 2, 3, … Adding 0 gives the whole numbers, and adding the negatives of the natural numbers gives the integers. Every natural number is a whole number and every whole number is an integer, but 0 is a whole number that is not natural.
A rational number can be written as p/q with integers p and q and q ≠ 0. Its decimal form either terminates or is non-terminating and repeating. An irrational number has a decimal form that neither terminates nor repeats, for example √2, √3 and π. Rational and irrational numbers together make up the real numbers.
Sum, difference or product of a rational number and an irrational number (non-zero in the product case) is irrational. Note that 22/7 is rational and is only an approximation of π, which is irrational.
Numbers whose square is negative are imaginary (like 3i, with i = √−1). A real number plus an imaginary number gives a complex number.

Primes, Co-primes and Composite Numbers

A prime number has exactly two factors, 1 and itself. There are 25 primes below 100, 2 is the only even prime, and 1 is neither prime nor composite. Composite numbers are natural numbers greater than 1 that are not prime.
Test for primality: find the smallest n with n² greater than the number, then test divisibility by every prime below n. If none divides it, the number is prime.
Two numbers with no common factor other than 1 are co-prime (they need not be prime themselves). Primes differing by 2, like (11, 13), are twin primes. Every composite number can be written in exactly one way as a product of primes.

Illustration 1. Is 571 a prime number?
Answer:
23² = 529 and 24² = 576, so test primes up to 23. Check 2, 3, 5, 7, 11, 13, 17, 19 and 23: 571 = 7 × 81 + 4 = 11 × 51 + 10 = 13 × 43 + 12 = 17 × 33 + 10 = 19 × 30 + 1 = 23 × 24 + 19 and the digit sum 13 rules out 3.
None divides 571, so 571 is prime.

Recurring Decimals as Fractions

A pure recurring decimal: write the repeating block once on top and put as many 9s below as there are digits in the block. For example 0.\(\overline{6}\) = 6/9 = 2/3 and 0.\(\overline{18}\) = 18/99 = 2/11.
A mixed recurring decimal such as 2.74\(\overline{35}\): take the whole digit string, subtract the non-repeating part, and divide by as many 9s as repeating digits followed by as many 0s as non-repeating digits after the point.
A fraction in lowest terms is a terminating decimal exactly when its denominator has no prime factors other than 2 and 5.

Illustration 2. Convert 0.\(\overline{432}\) and 0.4\(\overline{35}\) into fractions.
Answer:
0.\(\overline{432}\) = 432/999 = 16/37.
0.4\(\overline{35}\) = (435 − 4)/990 = 431/990.
16/37 and 431/990.

Squares and Cubes

A perfect square cannot end in 2, 3, 7 or 8, and the number of zeros at its end is always even. The square of a natural number n equals the sum of the first n odd numbers, and (n + 1)² − n² = 2n + 1. Every perfect square is a multiple of 3 or one more than a multiple of 3, and a multiple of 4 or one more than a multiple of 4.
Squares ending in 5: for A5, the square is A(A + 1) followed by 25. For example 85² = 7225 because 8 × 9 = 72. For numbers near a round number use x² = (x + y)(x − y) + y², for example 102² = 104 × 100 + 4 = 10404.
Cube roots of perfect cubes: the last digit of the cube root is found from the last digit of the cube (1↔1, 4↔4, 5↔5, 6↔6, 9↔9, 0↔0, 2↔8, 3↔7). The cube of a multiple of 3 is a multiple of 27.

Quick Multiplication

To multiply by 99, 999, … attach zeros to the number and subtract the number: 3893 × 99 = 389300 − 3893 = 385407. To multiply by 11, 101, 1001, … attach zeros and add the number: 4782 × 11 = 47820 + 4782 = 52602.
To multiply by 15, 25 or 35, double the multiplier, multiply, then halve: 7054 × 15 = 211620 ÷ 2 = 105810. For powers of 5 (5, 25, 125, 625) attach zeros and divide by the matching power of 2: 4739 × 25 = 473900 ÷ 4 = 118475.

Divisibility Tests

By 2: last digit even. By 3 or 9: the digit sum is divisible by 3 or 9. By 4: the last two digits form a number divisible by 4. By 5: last digit 0 or 5. By 6: even and divisible by 3. By 8: last three digits divisible by 8. By 10: ends in 0. By 12: divisible by both 3 and 4. By 25 or 125: last two or three digits divisible by 25 or 125.
By 7: double the last digit and subtract it from the rest; repeat if needed. By 11: the difference between the sum of digits at odd places and at even places is 0 or a multiple of 11. By 88: divisible by both 8 and 11.

Illustration 3. Is 6853 divisible by 7? Is 91728 divisible by 11?
Answer:
6853: 685 − 2 × 3 = 679; 67 − 2 × 9 = 49, which is divisible by 7, so 6853 is divisible by 7 (6853 = 7 × 979).
91728: digits at odd places 9 + 7 + 8 = 24, at even places 1 + 2 = 3; difference 21 is not divisible by 11, so 91728 is not divisible by 11.

HCF and LCM: Standard Problem Types

HCF × LCM = product of the two numbers. Greatest number dividing x, y, z exactly = HCF(x, y, z); least number divisible by x, y, z = LCM(x, y, z).
Greatest number dividing x, y, z leaving remainders a, b, c = HCF(x − a, y − b, z − c). If the remainder is the same but unknown, take the HCF of the differences of the numbers.
Least number leaving remainders a, b, c on division by x, y, z: if x − a = y − b = z − c = k, the number is LCM − k. If the remainder is the same r each time, it is LCM + r.
Greatest n-digit number with a given property: subtract from the greatest n-digit number its remainder on division by the LCM, then add any required remainder. For the smallest n-digit number, add (LCM − remainder) to the smallest n-digit number and then add any required remainder.
Number of multiples of m up to N is the quotient of N ÷ m. If a number leaves remainders r₁ and r₂ when divided successively by d₁ and d₂, its remainder on division by d₁d₂ is d₁r₂ + r₁.

Illustration 4. Find the greatest number that divides 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.
Answer:
Subtract the remainders: 144, 240, 612. HCF(144, 240, 612) = 12.
12.

Illustration 5. Find the least number which, when divided by 36, 48 and 64, leaves remainders 25, 37 and 53.
Answer:
36 − 25 = 48 − 37 = 64 − 53 = 11, and LCM(36, 48, 64) = 576.
Required number = 576 − 11 = 565.

Illustration 6. Find the greatest number of four digits which, when divided by 12, 18, 21 and 28, leaves 3 as remainder in each case.
Answer:
LCM = 252. 9999 ÷ 252 leaves remainder 171.
Required number = 9999 − 171 + 3 = 9831.

Illustration 7. The HCF and LCM of two numbers are 63 and 1260. If one of the numbers is 315, find the other.
Answer:
Other number = (63 × 1260) ÷ 315 = 252.

Counting Factors

If N = aᵖ × bᵠ × cʳ × … with a, b, c prime, then the number of factors is (p + 1)(q + 1)(r + 1)…, and the sum of the factors is [(aᵖ⁺¹ − 1)/(a − 1)] × [(bᵠ⁺¹ − 1)/(b − 1)] × …
The product of all the factors of N is \( N^{n/2} \), where n is the number of factors. The number of odd factors uses only the odd prime powers; the number of even factors = total factors − odd factors.
The number of ways of writing N as a product of two factors is half the number of factors (when N is not a perfect square). For a perfect square it is (number of factors + 1)/2, and (number of factors − 1)/2 if the two factors must be distinct.

Illustration 8. Find the number of factors and the sum of the factors of 270.
Answer:
270 = 2 × 3³ × 5. Number of factors = 2 × 4 × 2 = 16.
Sum = (2² − 1)/(2 − 1) × (3⁴ − 1)/(3 − 1) × (5² − 1)/(5 − 1) = 3 × 40 × 6 = 720.
16 factors; sum 720.

Illustration 9. In how many ways can 576 be written as a product of two distinct factors?
Answer:
576 = 2⁶ × 3², so the number of factors = 7 × 3 = 21.
Distinct pairs = (21 − 1)/2 = 10.

Illustration 10. Find the number of odd factors and even factors of 180.
Answer:
180 = 2² × 3² × 5. Total factors = 3 × 3 × 2 = 18. Odd factors come from 3² × 5: 3 × 2 = 6.
Even factors = 18 − 6 = 12.