Numbers in general form

Any two digit number xy made of digits x and y can be written as = 10 X x + 1X y

Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz

EXAMPLE 1: Write the number 12 in generalized form.

SOLUTION: 12 = 10X1 + 1X2 = 10+2

EXAMPLE 2: Write the number 10X3 + 4 in usual form.

SOLUTION: 10X3 + 4 = 30 + 4 = 34

Games with numbers

Reversing a two-digit number:
STEP 1: Suppose we choose a number xy, which is a short form for the 2-digit number 10x + y.
STEP 2: On reversing the digits, we get the number yx = 10y + x.
STEP 3: If we add the two numbers we get: (10x + y) + (10y + x) = 11x + 11y = 11 (x + y).
So, the sum is always a multiple of 11.
Observe here that if we divide the sum by 11, the quotient is x + y, which is exactly the sum of the digits of chosen number xy.

Reversing a 3-digit number:
STEP 1: Let the 3-digit number be xyz = 100x + 10y + z.
STEP 2: After reversing the order of the digits, we get the number zyx= 100z + 10y + x.
STEP 3: On subtraction:
If x > z, then the difference between the numbers is (100x + 10y + z) – (100z + 10y + x) = 100x + 10y + z – 100z – 10y – x = 99x – 99z = 99(x – z).
• If z>a, then the difference between the numbers is (100z + 10y + x) – (100x + 10y + z) = 99z – 99x = 99(z – x).
• And, of course, if x = z, the difference is 0. In each case, the resulting number is divisible by 99. So the remainder is 0.
Observe that quotient is x – z or z – x.

Forming three digit numbers with given 3-digits:

xyz = 100x + 10y + z
zxy = 100z + 10x + y
yzx = 100y + 10z + x
xyz + zxy + yzx = 111(x + y + z)
= 37 × 3(x + y + z), which is divisible by 37

EXAMPLE 1: Prove that the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.

SOLUTION: 12= 10X1 + 2
21= 10X2 + 1
12+21= 11X3= 11X(1+2) [PROVED]

Hence, the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.

EXAMPLE 2: Form 3-digit numbers with given 3 digit no. 123

SOLUTION: Combinations of the number 123 are 231,312 [Keeping in mind the no.s are in order]

123+231+312= 666= (1+2+3)X111= (1+2+3)X3X37

Hence, a 3 digit number formed by another 3-digit number can always be divided by 37.

Letter for digits

Here letters take the place of digits in an arithmetic ‘addition’, and the we have to find out which letter represents which digit.

Each letter must stand for just one digit. Each digit must be represented by just one letter.

The first digit of a number can’t be zero. Hence, we write the no. ‘ninety-five’ as 95, and not as 095, or 0095.

EXAMPLE 1: Find X and Y in the addition. X + X + X
Y X

SOLUTION: This has two letters X and Y whose values are to be found.
Study the addition in the one's column: the sum of three X’s is a number whose ones digit is X.
Therefore, the sum of two X’s must be a number whose ones digit is 0.
This happens only for X = 0 and X = 5.
If X = 0, then the sum is 0 + 0 + 0 = 0, which makes Y = 0 too.
We do not want this (as it makes X = Y, and then the tens digit of YX too becomes 0), so we reject this possibility.
So, X = 5.
Therefore, the puzzle is solved as shown below.

5
+ 5
+ 5
15 That is, X = 5 and Y = 1.

EXAMPLE 2: Find the digits X and Y.
Y X
× Y 3
5 7 X

SOLUTION: This also has two letters X and Y whose values are to be found.
Since the ones digit of 3 × X is X, it must be that X = 0 or X = 5.
Now, look at Y.
If Y = 1, then YX × Y3 would at most be equal to 19 × 19; that is, it would at most be equal to 361.
But the product here is 57X, which is more than 500.
So we cannot have Y = 1.
If Y = 3, then YX × Y3 would be more than 30 × 30; that is, more than 900.
But 57X is less than 600.
So, Y cannot be equal to 3.
Hence, Y = 2 .
Therefore, the multiplication is either 20 × 23 or 25 × 23.
But, 20 × 23 = 460. Thus, the second one is the correct multiplication.
The second one works out correctly, 25 × 23 = 575. The answer is X = 5, Y = 2.
2 5
×2 3
575

Tests of Divisibility

Divisibility by 10: If the ones digit of a number is 0, then the number is a multiple of 10; and if the ones digit is not 0, then the number is not a multiple of 10. So, we get a test of divisibility by 10.

Divisibility by 5: If the ones digit of a number is 0 or 5, then it is divisible by 5.

Divisibility by 2: If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

Divisibility by 9 and 3:
(a) A number is divisible by 9 if the sum of its digits is divisible by 9.
(b) A number X is divisible by 3 if the sum of its digits is divisible by 3. Otherwise, it is not divisible by 3.

EXAMPLE 1: Check the divisibility of 15284 by 3.

SOLUTION: The sum of the digits of 15284 is 1 + 5 + 2 + 8 + 4 = 20. This number is not divisible by 3. We conclude that 15284 too is not divisible by 3.

EXAMPLE 2: If the three-digit number 12a is divisible by 9, what is the value of a?

SOLUTION: Since 12a is divisible by 9, sum of its digits, i.e., 1 + 2 + a should be divisible by 9, i.e., 3 + a should be divisible by 9.
This is possible when 3 + a = 9 or 18, .... But, since a is a digit, therefore, 3 + a = 9, i.e., x = 6.

Practice these questions

Q1) If 12a3 is a multiple of 3, where a is a digit, what might be the values of a?

Q2) If 34a is a multiple of 3, where a is a digit, what is the value of a?

Q3) If 56a7 is a multiple of 9, where a is a digit, what is the value of a?

Q4) Find the value of the letters 1 2 X
+ 6 A Y
X 0 9

Q5) Find the value of the letters 2 A B
+A B 1
B 1 8

Q6) If 417 is added with its other combinations then check if the sum is divided by 37 or not.

Q7) Write the number 100 × x + 10 × y + z in its usual form.

Q8) Write the number 302 in its generalised form.

Recap

Numbers are four types: Natural Number, Whole Number, Integer, and Rational Numbers.
Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz
xyz = 100x + 10y + z
zxy = 100z + 10x + y
yzx = 100y + 10z + x
xyz + zxy + yzx = 111(x + y + z) = 37 × 3(x + y + z), which is divisible by 37
If the ones digit of a number is 0, then the number is a multiple of 10
If the digit at one’s place of a number is 0 or 5, then the number is divisible by 5.
If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

Quiz for Playing with Numbers