Rational Numbers | IT

Question 12

What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?

Solution

A number ending in zeros can be written as a product of another number and a power of 10.

Step 1 — Observing the pattern

Let us look at some examples of numbers ending in zeros and their squares. We will count the number of zeros at the end of each number.

$10^2 = 100$ $20^2 = 400$ $50^2 = 2500$ $100^2 = 10000$ $300^2 = 90000$ $1000^2 = 1000000$

$\boxed{\text{We notice that the number of zeros at the end of a square is double the number of zeros at the end of the original number.}}$

Diagram 1

Step 2 — Explaining the pattern

Let a number be $N$ . Suppose $N$ has $k$ zeros at its end. This means $N$ can be written as $M \times 10^k$ . Here, $M$ is a whole number that does not end in zero. For example, if $N = 300$ , then $k=2$ and $M=3$ . Now, let us find the square of $N$ .

$N^2 = (M \times 10^k)^2$

We use the exponent rule $(ab)^c = a^c b^c$ .

$N^2 = M^2 \times (10^k)^2$

We use another exponent rule $(a^b)^c = a^{bc}$ .

$N^2 = M^2 \times 10^{2k}$

Since $M$ does not end in zero, $M$ is not a multiple of 10. This means the prime factors of $M$ do not include both 2 and 5. Therefore, $M^2$ will also not end in zero. For example, if $M=3$ , $M^2=9$ . If $M=12$ , $M^2=144$ . So, the number of zeros at the end of $N^2$ is determined by the $10^{2k}$ part. The term $10^{2k}$ means there are $2k$ zeros. Since $k$ is the number of zeros in the original number, $2k$ is double the number of zeros.

$\boxed{\text{Yes, this will always happen.}}$

Step 3 — Concluding about perfect squares

From Step 2, we found that if a number has $k$ zeros at its end, its square will have $2k$ zeros at its end. The number $k$ can be any whole number (0, 1, 2, 3, ...). The number of zeros in the square, $2k$ , will always be an even number. For example, if $k=1$ , $2k=2$ (an even number). If $k=2$ , $2k=4$ (an even number). If $k=3$ , $2k=6$ (an even number). So, any perfect square that ends in zeros must have an even number of zeros.

$\boxed{\text{Consequently, we can say that perfect squares can only have an even number of zeros at the end.}}$

Answer

(i) We notice that the number of zeros at the end of a square is double the number of zeros at the end of the original number. (ii) Yes, this will always happen. (iii) Consequently, we can say that perfect squares can only have an even number of zeros at the end.

More questions in IT

Context: Queen Ratnamanjuri left a puzzle in her will for her son Khoisnam and 99 relatives. They are in a room with 100 lockers, numbered 1 to 100.

Person 1 opens every locker.
Person 2 toggles every 2nd locker (closes if open, opens if closed).
Person 3 toggles every 3rd locker (3rd, 6th, 9th, ...).
Person 4 toggles every 4th locker (4th, 8th, 12th, ...). This continues until all 100 get their turn.

Q. Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer?

Hint: Find out how many times each locker is toggled.

Does every number have an even number of factors?

Can you use this insight to find more numbers with an odd number of factors?

Context: In a room with 100 lockers, only lockers whose numbers are square numbers remain open.

Q. Write the locker numbers that remain open.

Find the squares of the first 30 natural numbers and fill in the table below.

Context: Patterns and Properties of Perfect Squares

Find the squares of the first 30 natural numbers and fill in the table below.

Q. What patterns do you notice? Share your observations and make conjectures.

If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?

Write 5 numbers such that you can determine by looking at their units digit that they are not squares.

Let us consider square numbers ending in 6: $16 = 4^2$ , $36 = 6^2$ , $196 = 14^2$ , $256 = 16^2$ , $576 = 24^2$ , and $676 = 26^2$ . Which of the following numbers have the digit 6 in the units place?

(i) $38^2$ (ii) $34^2$ (iii) $46^2$ (iv) $56^2$ (v) $74^2$ (vi) $82^2$

Q10

Find more such patterns by observing the numbers and their squares from the table you filled earlier.

Q11

If a number contains 3 zeros at the end, how many zeros will its square have at the end?

Q12

Q13

What can you say about the parity of a number and its square?

Q14

Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

Q15

How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?

Q16

Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

Q17

Find whether 1156 and 2800 are perfect squares using prime factorisation.

Q18

How many cubes of side 1 cm make a cube of side 2 cm?

Q19

How many cubes of side 1 cm will make a cube of side 3 cm?

Q20

Is 9 a cube?

Q21

Can you estimate the number of unit cubes in a cube with an edge length of 4 units?

Q22

Complete the table below.

Q23

What patterns do you notice in the table above?

Q24

We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?

Q25

Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?

Q26

Can a cube end with exactly two zeroes (00)? Explain.

Q27

The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.

Q28

Context: Look at the following pattern of consecutive odd numbers: $1 = 1 = 1^3$ $3 + 5 = 8 = 2^3$ $7 + 9 + 11 = 27 = 3^3$ $13 + 15 + 17 + 19 = 64 = 4^3$ $21 + 23 + 25 + 27 + 29 = 125 = 5^3$ $31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3$

Later in this series, we get the following set of consecutive numbers: $91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109$

Q. Can you tell what this sum is without doing the calculation?

Q29

Find the cube roots of these numbers:

(i) $\sqrt[3]{64} =$ (ii) $\sqrt[3]{512} =$ (iii) $\sqrt[3]{729} =$

Q30

Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?