Playing With Numbers: Puzzles, Number Games, Examples

Playing with Numbers involves several fun activities like number puzzles, arranging numbers, interchanging or reversing the digits, solving magic squares, etc. The magic square is a number puzzle. You might have solved magic squares. Do you still remember the rules for a magic square? All numbers in a row, column or diagonal must add up to the same number in a magic square.

In this article, we will discuss about Fun with numbers using a generalised form of numbers, number games and puzzles, even and odd numbers, tests of divisibility, etc. Scroll down to learn more!

Playing With Numbers: Fun With Numbers

Let us take any number and interchange or reverse the digits. Next, we will add the reversed numbers to the original number. Let’s do this activity for a few numbers.

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Did you notice any patterns? Is there any pattern in the sums?

Yes, for any two-digit number, if we reverse the digits and add to the original number, the sum is always a multiple of \(11\).

Will this hold true if we subtract instead of adding the numbers? Will there again be a pattern as well?

Let us check it out by taking other two-digit numbers.

Now, is there any pattern in the difference between the original number and the reversed number? If the patterns are valid for any two-digit number, then how can one explain what is going on?

For all these queries, we first have to learn to write a number in the generalised form.

Generalised Form of a Number

For any two-digit number, we know that the face value and place value of the unit’s place digit remains the same, but the place value of the ten’s place digit is ten times the face value of the digit. The value of the number is the sum of the individual place values of each digit.

For example, \(45 = 4 \times 10 + 5 \times 1\) or \(45 = 4 \times 10 + 5\)

In general, for any two-digit number, we can write the number in the generalized form as \(10x + y\), where \(x\) is the digit in the ten’s place, and \(y\) is the digit in the unit’s place in the number.

And, if we reverse the digits, the number becomes \(10y + x\). Thus, the sum of the original number and the reversed number \( = (10x + y) + (10y + x)\)
\( = 10x + y + 10y + x\)
\( = 10x + x + 10y + y\)
\( = 11x + 11y\)
\( = 11(x + y)\)

Hence, the sum will always be a multiple of \(11\).

What happens if we take the difference of three-digit numbers after reversing the digits? Will they follow the same patterns? Let us quickly check for three-digit numbers.

All the numbers in the difference column are multiples of \(99\). Let us see why this is so, using the generalized form of a number. We can write any three-digit number as \(100x + 10y + z\). Thus, the reversed number will be \(100z + 10y + x\).

Therefore, the difference of the two numbers \( = (100x + 10y + z) – (100z + 10y + x)\)
\( = 100x + 10y + z – 100z – 10y – x\)
\( = 100x – x + 10y – 10y + z – 100z\)
\( = 99x – 99z\)
\( = 99(x – z)\)

Hence, the difference will always be a multiple of \(99\).

Number Games and Puzzles

Now, let us have some fun using letters as numbers and solving the math puzzles. Here, we have puzzles in which letters take the place of digits in an arithmetic sum, and the problem is to find out which letter represents which digit. So, it’s like cracking a secret code. Here we stick to the properties of addition and multiplication.

Let us have a look at the rules which we follow for solving such types of puzzles.

1. Each letter in the puzzle represents one digit only.
2. The first digit of a number can never be \(0\).

Let us understand it better by taking a couple of examples.

Example 1: Find \(A\) in addition.
\(\begin{gathered} \,\,\,\,31\,A \hfill \\\underline { + 1A3}  \hfill \\\underline {\,\,\,501}  \hfill \\\end{gathered} \)
Solution: There is only one letter \(A\) whose value has to find out. Check out the addition in one’s column, i.e.,  \(A + 3\). The unit digit of the answer is \(1\). As \(3\) is greater than \(1\), the answer has to be \(11\) 
For this, digit \(A\) should be \(8\). So, we can write the puzzle as;
\(\begin{gathered}\,\,\,\,31\,8 \hfill \\\underline { + 183}  \hfill \\\underline {\,\,\,501}  \hfill \\\end{gathered} \)

Example 2: The letters in the given problem stand for the missing digit. Find the missing digits.
\(2A5 + B23 = 368\)
Solution: We can write the puzzle \(2A5 + B23 = 368\) as
\(\begin{gathered} \,\,\,\,2A5 \hfill \\\underline { + B23}  \hfill \\\underline {\,\,\,368}  \hfill \\\end{gathered} \)
The ten’s place digits are \(A\) and \(2\), and the ten’s place digit of the answer is \(6\). Thus, in \(A = 6 – 2 = 4\), as \(4 + 2 = 6\).
In the hundred’s place, \(2 + B = 3\), thus \(B = 1\)
So, we can write the puzzle as;
\(\begin{gathered} \,\,\,\,245 \hfill \\\underline { + 123}  \hfill \\\underline {\,\,\,368}  \hfill \\\end{gathered} \)

Even and Odd Numbers

All multiples of \(2\) are called even numbers. For example, \(2,\,4,\,6,\,8,\,10,\,12,\,14,\,16,\,18,\, \ldots \) … are all multiples of \(2\) and thus are even numbers. Every even number is divisible by \(2\) or we can say that \(2\) is a factor of every even number.

Numbers that are not divisible by \(2\) are called odd numbers. For example, \(1,\,3,5,\,7,\,9,\,11,\,13,\,15,\,17,\, \ldots \)… are not divisible by \(2\) and thus are known as odd numbers.

Every number is either even or odd. Also, a number cannot be both even or odd.

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Tests of Divisibility

We have learnt many rules and properties of integers, whole numbers, etc., in mathematics. Let us learn the divisibility rules of a number in detail.

1. Divisibility by 2

We know that all even numbers are divisible by \(2\) and all odd numbers will have a remainder of \(1\) if we divide them by \(2\). In simple words, a number is divisible by \(2\) if its unit digit is \({\rm{0,}}\,{\rm{2,}}\,{\rm{4,}}\,{\rm{6}}\) and \({\rm{8}}\), and if the number has either \({\rm{1,}}\,{\rm{3,}}\,{\rm{5,}}\,{\rm{7}}\) and \(9\), in its unit’s place, then the number is not divisible by \(2\).

For example, \(6,\,118,\,200,\,1112,\,1554\) are all divisible by \(2\).

2. Divisibility by 3

A number is divisible by \(3\) if the sum of its digits is divisible by \(3\). For example, consider the number \(4219\).
Thus, the sum of its digits \( = 4 + 2 + 1 + 9 = 16\), which is not divisible by \(3\).

Again, consider the number \(5193\), the sum of its digits \( = 5 + 1 + 9 + 3 = 18\), which is divisible by \(3\).

3. Divisibility by 4

A number is divisible by \(4\), if the number formed by its digit in ten’s place and unit’s place is divisible by \(4\). In \(432\), the number formed by its digits in ten’s place and unit’s place is \(32\) and \(32\) is divisible by \(4\).

4. Divisibility by 5

A number is divisible by \(5\) if its unit digit is either \(0\) or \(10\). For example, numbers like \(120,\,115,\,2020,\,100,\,5900\), etc., are all divisible by \(5\).

5. Divisibility by 6

A number is divisible by \(6\) if it is divisible by \(2\) and \(3\). For example, consider the number \(88536\). The number is divisible by \(2\) as the digit at its unit’s place is \(6\) which is even and the sum of the digits \( = 8 + 8 + 5 + 3 + 6 = 30\), divisible by \(3\).

Hence, the number \(88536\) is divisible by \(6\).

6. Divisibility by 7

The difference between twice the unit digit of the given number and the remaining part of the given number should be a multiple of \(7\) or it should be equal to \(0\), then we say a number is divisible by \(7\). For example, \(133\), double the last digit of the number \(3\) is \(6\) Now, \(13 – 6 = 7\). Since \(7\) is divisible by \(7\).
Hence, \(133\) is also divisible by \(7\).

7. Divisibility by 8

A number is divisible by \(8\) if the number formed by its digits at hundred’s place, ten’s place, and unit’s place is divisible by \(8\). For example, in \(9408\) the number formed by its digit’s in hundred’s place, ten’s place and unit’s place is \(408\), and \(408\) is divisible by \(8\). Hence, \(9408\) is divisible by \(8\).

8. Divisibility by 9

A number is divisible by \(9\) if the sum of its digits is divisible by \(9\). For example, consider the number \(3215\).
Thus, the sum of its digits \( = 3 + 2 + 1 + 5 = 11\), which is not divisible by \(9\).

Again, consider the number \(9693\), the sum of its digits\( = 9 + 6 + 9 + 3 = 27\), which is divisible by \(9\).

9. Divisibility by 10

A number is divisible by \(10\), if its unit digit is \(0\). For example, numbers like \(100,\,1110,\,1320\) etc., are all divisible by \(10\).

Solved Examples on Playing With Numbers

Q.1. Express \(975\) in the generalised form.
Ans: We can write any three-digit number as \(100x + 10y + z\) in generalized form. Thus, \(975 = 9 \times 100 + 7 \times 10 + 5\)

Q.2. The letters stand for the missing digits. Find the missing digits.
\(32D \times 7 = 2275\)
Ans: We can write \(32D \times 7 = 2275\) as
\(\begin{gathered} \,\,\,\,32D \hfill \\\underline { \times \,\,\,\,\,\,\,\,7}  \hfill \\\underline {2275}  \hfill \\\end{gathered} \)
The unit’s place digit in the answer is \(5.7 \times 5 = 35\). Thus, \(D\) must be \(5\). Let us verify the rest of the multiplication with \(D\) as \(5\).
Therefore, \(325 \times 7 = 2275\), which matches the given number.
Hence, \(D = 5\).

Q.3. Check whether the following numbers are divisible by \(2\) and \(3\).
a) \(362\)
b) \(111\)
Ans: A number is divisible by \(2\) if its unit digit is \(0,\,2,\,4,\,6\) and \(8\), whereas a number is divisible by \(3\) if the sum of its digits is divisible by \(3\).
a) In the number \(362\), the unit digit is \(2\), which is divisible by \(2\). Hence, the number is divisible by \(2\). Now, the sum of the digits \( = 3 + 6 + 2 = 11\), which is not divisible by \(3\). Thus, the number \(362\) is divisible by \(2\) but not by \(3\). 
b) In the number \(111\), the unit digit is \(1\), which is not divisible by \(2\). Hence, the number is not divisible by \(2\). Now, the sum of the digits \( = 1 + 1 + 1 = 3\), which is divisible by \(3\). Thus, the number \(362\) is not divisible by \(2\) but divisible by \(3\).

Q.4. There are three two-digit numbers: \({\rm{46,}}\,{\rm{72}}\) and \(19\).  Show that in each case, if the digits are reversed, and the difference between the original number and the new number is calculated, the answer is a multiple of \(9\).
Ans:

All the numbers in the difference column are multiples of \(9\).
We can write any two-digit number as \(10y + x\). Thus, the reversed number will be \(10x + y\).
Therefore, the difference of the two numbers \( = (10y + x) – (10x + y)\)
\( = 10y + x – 10x – y\)
\( = 10y – y + x – 10x\)
\( = 9y – 9x\)
\( = 9(y – x)\)

Q.5. Check whether the following numbers are divisible by \(5\).
a) (2015\)
b) \(1117\)
Ans: A number is divisible by \(5\) if its unit digit is either \(0\) or \(5\).
a) In \(2015\), the digit at the unit’s place is \(5\), and thus, the number is divisible by \(5\).
b) In \(1117\), the digit at the unit’s place is \(7\), and thus, the number is not divisible by \(7\).

Summary

In this article, we learned to find the generalised form of a two-digit and a three-digit number. One should note that for any two-digit number, the face value and place value of the unit’s place digit remains the same. However, the place value of the ten’s place digit is ten times the face value of the digit. We also learned that all multiples of 2 are referred to as even numbers. Furthermore, numbers that are not divisible by 2 are referred to as odd numbers. In this article, we also learned the tests of divisibility of 2, 3, 4, 6, 7, 8, 9 and 10.

FAQs on Playing With Numbers

Q.1. Determine the smallest three-digit number, which is exactly divisible by 6, 8 and 12.
Ans: To find the smallest three-digit number, we first need to find the LCM of the given numbers. Thus, the LCM of \(6,\,8\) and \(12 = 24\). We can see that \(24 \times 4 = 96\)and \(24 \times 5 = 120\).
Thus, the smallest \(3\)digit number, which is exactly divisible by \({\rm{6,}}\,{\rm{8}}\) and \(12\) is \(120\).

Q.2. Find the prime factorisation of 225.
Ans: The prime factors of \(225 = 3 \times 3 \times 5 \times 5 = {3^2} \times {5^2}\)

Q.3. Define factor and multiple.
Ans:  We can define a multiple as a number divided by another number a certain number of times without a remainder. At the same time, a factor is one of two or more numbers that divide the given number without a remainder. In simple words, multiples are the product of two factors.

Q.4. List out numbers between 1 – 100 having precisely 3 factors.
Ans: The numbers between \(1\) and \(100\) with exactly three factors are \({\rm{4,}}\,{\rm{9,}}\,{\rm{25}}\) and \({\rm{49}}\)
Factors of \(4\) are \({\rm{1,}}\,{\rm{2}}\) and \(4\) factors of \(9\) are \({\rm{1,}}\,{\rm{3}}\) and \(9\), factors of \(25\) are \({\rm{1,}}\,{\rm{5}}\) and \(25\), and factors of \(49\) are \({\rm{1,}}\,{\rm{7}}\) and \(49\).

Q.5. Determine even and odd numbers:
13, 28, 41, 99, 50, 255, 132, 664, 59, 29, 71, 165, 14
Ans: Even numbers are \(28,\,50,\,132,\,664,\,14\).
Odd numbers are \(13,\,41,\,99,\,255,\,59,\,29,\,71,\,165\).

We hope this detailed article on playing with numbers helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!