Playing with Numbers

Representing numbers in general form

Representing two-digit numbers in their general form

Consider the two-digit number 89.

Recalling the concept of 'Place value in Numbers' from our lower classes, we know that 89 consists of 8 TENS and 9 ONES.

Therefore, 89 can also be represented as (10×8)+9.

From the example discussed previously, it is understood that two-digit numbers will be formed using some TENS and some ONES.

Let the some number of TENS be 'a' and the some number of ONES be 'b'.

Two-digit number 'ab' can also be represented as (10×a)+b or 10a+b.

Representing three-digit numbers in their general form

Let us consider a three-digit number 273.

Similar to the construction of two-digit numbers, three-digit numbers are formed using HUNDREDS, TENS and ONES.

In this case, let HUNDREDS be represented as 'a', TENS be represented as 'b', and ONES be represented as 'c'.

Therefore, the general form of the three-digit number 'abc' is (100×a)+(10×b)+c or 100a+10b+c.

Therefore, 273 becomes (100×2)+(10×7)+3.

Reversing the two digit numbers

Logic 1: The sum of the two numbers is always a multiple of 11

The general form of the two-digit number 'ab' is (10×a)+b.

We know that to reverse a number, we should swap the numbers in TENS and ONES position.

Therefore, the general form of its reverse 'ba' is (10×b)+a.

Let us try to find the sum of both these numbers.

ab+ba=[(10×a)+b]+[(10×b)+a]

=10a+b+10b+a

=11a+11b

=11(a+b)

Therefore, the sum of two-digit numbers and their reverse is always a multiple of 11.

Logic 2: The difference between the two numbers is always a multiple of 9

Let us now find the difference between the two numbers.

ab−ba=[(10×a)+b]−[(10×b)+a]

=10a+b−10b−a

=9a−9b

=9(a−b)

Thus, it is understood that the difference between the two numbers is a multiple of 9.

Example:

Let us try to apply both the logics discussed above on the number 91.

Reverse of the number =19

The sum of both these numbers =91+19

=110

=11×10

Therefore, by logic 1, it is proved that the sum of two-digit numbers and their reverse is a multiple of 11.

Difference between both these numbers =91−19

=72 =9×8

Therefore, by logic 2, it is proved that the difference between two-digit numbers and their reverse is a multiple of 9.

Tricks dealing with three-digit numbers

Trick 1: What happens to the difference between three-digit numbers and their reverse?

The difference between three digit numbers and their reverse is always a multiple of 99.

Consider the three-digit number 'abc'.

The general form of 'abc' is (100×a)+(10×b)+c or 100a+10b+c.

The reverse of 'abc' is 'cba'.

The general form of 'cba' is (100×c)+(10×b)+a or 100c+10b+a..

Now, let us find the difference between 'abc' and 'cba'.

If a>c:

abc−cba=(100a+10b+c)−(100c+10b+a)

=100a+10b+c−100c−10b−a

=99a−99c

=99(a−c)

If c>a:

cba−abc=(100c+10b+a)−(100a+10b+c)

=100c+10b+a−100c−10b−a

=99c−99a

=99(c−a)

Example:

Let us impose this logic on 325.

The reverse of 325 is 523.

Hence, to find the difference, we should subtract 325 from 523.

523−325=198 =99×2

Hence, it is proved that the difference between three-digit numbers and their reverse is a multiple of 99.

Trick 2: What happens when 3 forms of a three-digit number is summed up?

Consider the number 'abc'.

Form 1: abc=100a+10b+c(1)

To find the other forms of the number, shift the ONES digit to the number's left end.

Therefore, the number 'abc' becomes 'cab'.

The digit 'c' is shifted to the left end of the number.

Form 2: cab=100c+10a+b(2)

To form the third number, shift 'b' from 'cab' to the left end.

Form 3: bca=100b+10c+a(3)

On adding (1), (2), and (3), we get:

abc+cab+bca=100c+10a+b+100b+10c+a+100b+10c+a

=111(a+b+c)

=37×3(a+b+c)

Therefore, the sum of 3 forms of a three-digit number is always a multiple of 37.

Example:

Let us consider the number 128.

Form 1: 128

Form 2: 812

Form 3: 281

On adding the 3 numbers, we get:

128+812+281=1221

=37×33

Thus, the sum of three forms of a three-digit number is always a multiple of 37

Replace letters with digits to crack the code

1. Each of the letters of the alphabet will only constitute one-digit. 

2. The first digit of a number can never be '0'.

Let us try to solve the following puzzle.

In this puzzle, we should try to find the value of A.

Let us consider the entries in the ONES column.

A+4=1

It is now understood that A, when added with 4, gives a number ending with 1.

What are the numbers that end with 1?

1, 11, 21, 31,...

Let us look at the trial and error method:

When we consider A as 7, the final value is a number ending with 1.

Let us consider the HUNDREDS column.

1+A=8

Let us check if the condition A=7 is satisfied or not.

1+7=8

The condition is thus satisfied.

If we substitute 7 in the place of A, we get the following.

This type of puzzle solving is called 'Cracking of codes' or 'Cryptarithms'.

Introduction to divisibility tests

Divisibility test by 10

Let us consider a few multiples of 10.

50, 100, 200, 300,…

We can see that all these numbers end with 0, and it is the vital similarity between all multiples of 10.

Let us also observe a few other numbers, such as 13, 89, 155, etc. We find that all these numbers do not end with 0. Hence, these numbers are not divisible by 10.

Now let us consider the general form of a three-digit number 'abc'.

100a+10b+c

100a is a multiple of 10 as 100 itself is a multiple of 10. Any number multiplied by 100 automatically becomes a multiple of 10.

Similarly, 10b is also a multiple of 10.

To make the three-digit number 'abc' a multiple of 10, the digit 'c' plays a crucial role.

The three-digit number 'abc' will be a multiple of 10, only if 'c' is 0.

In other words, a number is divisible by 10 only if it ends with a 0.

Divisibility by 5

The numbers given below are all multiples of 5.

5, 10, 15, 20, 25, 30, 35, 40,…

Now, when we observe the multiples carefully, we find that the numbers always end with a 5 or a 0.

We can conclude that if a number ends with a 5 or a 0, it is divisible by 5.

In general form, the three-digit number 'abc' is 100a+10b+c.

From the divisibility test by 10, we understood that both 100a and 10b are multiples of 10.

10=2×5

Thus, 100a and 10b will also become a multiple of 5.

Hence, 'c' should either be 5 or 0 to make the number 'abc' a multiple of 5.

Divisibility by 2

Let us look at a continuous range of even numbers given below.

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28,…

After careful observation, we can understand that even numbers end only in 2, 4, 6, 8, and 0.

Consider any three-digit number 'abc' and its general form 100a+10b+c.

Here, 100a and 10b are multiples of 2 as 100 and 10 are multiples of 2 themselves.

Hence, 'c' can hold any value 2, 4, 6, 8, and 0 to make 'abc' divisible by 2.

Some other divisibility tests

Divisibility by 3 or 9

Consider the number 4239

The expanded form is (4×1000)+(2×100)+(3×10)+9.

=[4×(999+1)]+[2×(99+1)+[3×(9+1)]+9

=(4×999)+(4×1)+(2×99)+(2×1)+(3×9)+(3×1)+(9×1)

=(4×999)+(2×99)+(3×9)+(4+2+3+9)

Now, 4+2+3+9=18.

18 is both divisible by 3 and by 9. Therefore, 4239 is also divisible by 3 or 9.

Similarly, consider the number 2148

The expanded form is (2×1000)+(1×100+(4×10)+8.

=[2×(999+1)]+[1×(99+1)+[4×(9+1)]+8

=(2×999)+(2×1)+(1×99)+(1×1)+(4×9)+(4×1)+(8×1)

=(2×999)+(1×99)+(4×9)+(2+1+4+8)

Now, 2+1+4+8=15.

15 is divisible by 3 but not by 9. Therefore, 2148 is also divisible by 3 but not by 9.

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