JEE Mathematics Factors Multiples and Square Roots Notes Set 01

Factors and Multiples

A number a is a factor of b if b divides exactly by a, leaving remainder zero. For example, 3 and 5 are factors of 12 and 25 respectively, and 3 divides 12 four times exactly.
A multiple of a number is any number that the first divides exactly. 36 is a multiple of 2, 3, 4, 9 and 12.

Even and Odd Numbers

Multiples of 2 are even numbers (2, 4, 6, 8, ...); numbers that are not multiples of 2 are odd. In general 2n is even, while 2n + 1 and 2n − 1 are odd, for a whole number n.
Rules for sums and products
1. Sum or difference of two even numbers is even. Example: 18 + 22 = 40, 40 − 24 = 16.
2. Sum or difference of two odd numbers is even. Example: 25 + 31 = 56, 45 − 29 = 16.
3. Sum or difference of an even and an odd number is odd.
4. Product of two even numbers is even. Example: 8 × 4 = 32.
5. Product of an even and an odd number is even. Example: 9 × 8 = 72.
6. Product of two odd numbers is odd. Example: 5 × 9 = 45.
7. Any power of an even number is even (\( 2^4 = 16 \), \( 4^3 = 64 \)) and any power of an odd number is odd (\( 3^2 = 9 \), \( 5^3 = 125 \)).

Consecutive Numbers

Numbers in which each differs from the one before by 1 are consecutive: 5, 6, 7, 8, ... Consecutive odd numbers are 5, 7, 9, 11, ... and consecutive even numbers are 6, 8, 10, 12, ... (each pair differs by 2).

Square Roots

Since \( 2 \times 2 = 4 \) and also \( (-2) \times (-2) = 4 \), the number 4 is the square of both 2 and −2. The symbol \( \sqrt{\ } \) stands for the principal (positive) square root, so \( \sqrt{4} = 2 \), while the two square roots of 4 are \( \pm 2 \). Also \( \sqrt{0} = 0 \).
Quick tests for perfect squares. A perfect square can end only in 0, 1, 4, 5, 6 or 9. A number ending in 2, 3, 7 or 8 is never a perfect square. In its prime factorisation, every prime of a perfect square appears an even number of times.

Square Root by Prime Factorisation

Illustration 1. Find the square root of 27225 by factors.
Answer:
Divide repeatedly: 27225 = 5 × 5445, 5445 = 5 × 1089, 1089 = 3 × 363, 363 = 3 × 121, 121 = 11 × 11.
So \( 27225 = 5^2 \times 3^2 \times 11^2 \).
Take one from each pair: \( \sqrt{27225} = 5 \times 3 \times 11 = 165 \).

Illustration 2. Is 3528 a perfect square?
Answer:
\( 3528 = 2^3 \times 3^2 \times 7^2 \).
The prime 2 occurs three times, an odd number, so 2 is left without a partner. Hence 3528 is not a perfect square.
The last digit test agrees: 3528 ends in 8, and no perfect square ends in 8.

Illustration 3. By what least number must 19845 be multiplied to make a perfect square?
Answer:
\( 19845 = 3^4 \times 7^2 \times 5 \).
Only 5 occurs an odd number of times, so the least multiplier is 5. (The product is \( 99225 = 315^2 \).)

Illustration 4. Find a perfect square between 1000 and 2000 which is divisible by 13.
Answer:
A square divisible by 13 must contain \( 13^2 = 169 \) as a factor, so it has the form \( 169m^2 \).
\( 169 \times 4 = 676 \) is below 1000 and \( 169 \times 16 = 2704 \) is above 2000. Only \( 169 \times 9 = 1521 \) lies in the range.
The number is 1521, which equals \( (13 \times 3)^2 = 39^2 \).

Square Root by Estimation and Trial

Illustration 5. Find the square root of 5329.
Answer:
Since \( 70^2 = 4900 \) and \( 80^2 = 6400 \), the root is between 70 and 80. Write it as 70 + c.
\( (70 + c)^2 = 4900 + 140c + c^2 = 5329 \), so \( c(140 + c) = 429 \).
The number ends in 9, so the root ends in 3 or 7. Try c = 3: \( 3 \times 143 = 429 \). It works.
Hence \( \sqrt{5329} = 73 \).

Long Division Method

1. Mark the digits in pairs starting from the decimal point, going left for the whole part and right for the decimal part (a lone digit on the left forms the first group).
2. Find the largest number whose square is not more than the first group. Write it as the first digit of the root and subtract its square.
3. Bring down the next pair beside the remainder. Double the root so far and write it as the start of a new divisor.
4. Choose the largest digit d such that (divisor followed by d) × d is not more than the current number. Put d in the root and in the divisor, subtract, and repeat from step 3.
5. In a decimal, put the decimal point in the root as soon as the first decimal pair is brought down. A decimal number must have an even number of decimal places for its square root to terminate nicely, which is why pairs are made on the right (\( 0.14^2 = 0.0196 \), \( 2.5^2 = 6.25 \)).

Illustration 6. Find the square root of 47961.
Answer:
Pairs: 4 | 79 | 61. The largest square not more than 4 is \( 2^2 \), so the first digit is 2. Remainder 0; bring down 79.
Double 2 to get 4. Choose d with (40 + d) × d ≤ 79: d = 1 gives \( 41 \times 1 = 41 \). Remainder 38; bring down 61 to get 3861.
Double 21 to get 42. Choose d with (420 + d) × d ≤ 3861: d = 9 gives \( 429 \times 9 = 3861 \). Remainder 0.
So \( \sqrt{47961} = 219 \).

Illustration 7. Find the square root of 1218.7081.
Answer:
Pairs: 12 | 18 . 70 | 81. The largest square not more than 12 is \( 3^2 = 9 \). Remainder 3; bring down 18 to get 318.
Double 3 to get 6. d = 4 gives \( 64 \times 4 = 256 \). Remainder 62; bring down 70 to get 6270 (the decimal point now enters the root).
Double 34 to get 68. d = 9 gives \( 689 \times 9 = 6201 \). Remainder 69; bring down 81 to get 6981.
Double 349 to get 698. d = 1 gives \( 6981 \times 1 = 6981 \). Remainder 0.
So \( \sqrt{1218.7081} = 34.91 \).

Illustration 8. Find the fourth root of 108243216.
Answer:
The fourth root is the square root of the square root. First, \( \sqrt{108243216} = 10404 \) by long division (pairs 1 | 08 | 24 | 32 | 16).
Next, \( \sqrt{10404} = 102 \) (pairs 1 | 04 | 04).
So the fourth root of 108243216 is 102, since \( 102^4 = 108243216 \).

Square Roots of Non-Perfect Squares

To approximate \( \sqrt{2} \), write 2 as 2.00000000 and continue the long division, bringing down pairs of zeros. Carrying it out gives \( \sqrt{2} = 1.4142\ldots \), which is 1.414 to three decimal places. In the same way \( \sqrt{3} = 1.732\ldots \) (three decimal places) and \( \sqrt{5} = 2.2361 \) (four decimal places).
These roots are irrational: their decimal expansions never end and never repeat.

Two-Digit Number and Its Reverse

Illustration 9. The sum of a two-digit number and the number formed by reversing its digits is a perfect square. Find the numbers.
Answer:
Let the digits be x (tens) and y (units). The number is 10x + y and the reversed number is 10y + x.
Sum \( = 11x + 11y = 11(x + y) \). The largest possible x + y is 18.
For 11(x + y) to be a perfect square, 11 must divide it twice, so x + y must be a multiple of 11. Within 18, this means x + y = 11 and the sum is 121 = \( 11^2 \).
Digit pairs with sum 11 are (2, 9), (3, 8), (4, 7), (5, 6). The numbers are 29 and 92, 38 and 83, 47 and 74, 56 and 65. For example, 38 + 83 = 121.

Recurring Decimals as Fractions

Illustration 10. Express 3.3333... as a rational number.
Answer:
Let x = 3.3333... Then 10x = 33.3333...
Subtract: \( 10x - x = 30 \), so \( 9x = 30 \).
\( x = \dfrac{30}{9} = \dfrac{10}{3} = 3\dfrac{1}{3} \).

Illustration 11. Express \( 2.5\overline{73} \) in the form \( \dfrac{p}{q} \).
Answer:
Here x = 2.5737373..., where the block 73 repeats.
Multiply by 10 to move the non-repeating 5 before the point: 10x = 25.737373...
Multiply by 1000 to shift one full repeating block past it: 1000x = 2573.737373...
Subtract: \( 990x = 2548 \), so \( x = \dfrac{2548}{990} = \dfrac{1274}{495} \).