PLAYING WITH NUMBERS

(b) 3 Digit Number :

CONTENTS If number is abc or 100a + 10b + c, a  0 then the

 Generalised form of 2 Digit and reverse is cba or 100c + 10b + a

3 Digit numbers If a > c then

 Reversing the digits abc – cba = (100a + 10b + c) – (100c + 10b + a)

= 99 (a – c)
 Pythagorean triplets If c > a then

 Divisibility test cba – abc = (100c + 10b + a) – (100a + 10b + c)

= 99 (c – a)
That means difference of a 3 digit number and its
GENERALISED FORM OF 2 DIGIT AND
reverse number is divisible by 99.
3 DIGIT NUMBERS
* We can make more numbers from given no.
(a) 2 digit number has the tens place and the units
abc like bca, acb, bac, cab etc.
place
also abc + bca + cab = 111 (a + b + c)
Eg. 45 = 4 × 10 + 5, 93 = 9 × 10 + 3
= 37 × 3 (a + b + c)
Eg. ab = 10a + b, ba = 10b + a
 The number (abc + cab + bca) is divisible by
(b) 3 digit number has the hundreds place, the tens
37, 3 and a + b + c.
place and the units place.
eg. 927 :
Eg. 393 = 3 × 100 + 9 × 10 + 3
927 + 279 + 792 = 3 × 37 (9 + 2 + 7)
Eg. 492 = 4 × 100 + 9 × 10 + 2
= 3 × 37 × 18
Eg. 102 = 1 × 100 + 0 × 10 + 2
1998  3 = 666 = 18 × 37
Eg. abc = 100 a + 10 b + c
1998  37 = 54 = 18 × 3
Eg. cba = 100 c + 10 b + a
Eg. The usual form of 10 × 7 + 8 and 10 × 5 + 7 1998  18 = 111 = 3 × 37

are 78 and 57 respectively. Note : The first digit of a number can not be zero.
eg. 29 is a two digit number but 029 is not a 3
 REVERSING THE DIGITS
digit no.
(a) 2 Digit Number : If number is ab, a  0 then
 FIND THE DIGITS
reverse is ba. The difference of number & its
Ex.1 Find the value of x.
reverse is divisible by 9.
3 1 x
Eg. Reverse of 23 or 2×10 + 3 is 32 or 3×10 + 2 1 x 3
also 32 – 23 = 9 Its divisible by 9. 5 0 1
Sol. In ones column addition of x, 3 gives 1 Ex.5 Write a Pythagorean triplet whose smallest
 x may be 8
If x = 8 then we get a number whose ones digit is 1 member is 8.
& remaining 1 makes 2 + x in II column Sol. We can get Pythagorean triplet by using
 2 + 8 = 10
So 0 is tens digit of result and remaining 1 general from 2m, m2 – 1, m2 + 1.
makes 5 of sum of III column.
Let us first take m2 – 1 = 8
 x=8
So, m2 = 8 + 1 = 9
Ex.2 Find the value of x, y which gives m = 3
x
 x Therefore, 2m = 6
 x and m2 + 1 = 10
y x The triplet is thus 6, 8, 10. But 8 is not the
Sol. If x = 5 then 5 + 5 + 5 = 15
 y = 1, x = 5. smallest member of this.
So, let us try 2m = 8
Ex.3 Find the value of x, y
2 5 x 4 then m=4
 y 5 2 8 We get m2 – 1 = 16 – 1 = 15
1 2 1 0 2
Sol. x = 7, y = 9 and m2 + 1= 16 + 1 = 17
The triplet is 8, 15, 17 with 8 as the smallest
Ex.4 Find the value of x member.
2 x 7
7 2 x Ex.6 Find a Pythagorean triplet in which one
 x 7 2 member is 12.
1 x x 2
Sol. If we take m2 – 1 = 12
Sol. x=3
Then, m2 = 12 + 1 = 13
Then the value of m will not be an integer.
 PYTHAGOREAN TRIPLETS
So, we try to take m2 + 1 = 12.
If the square of a number is equal to sum of Again m2 = 11 will not give an integer value for m.
square other two numbers then these three So, let us try 2m = 12
numbers are called Pythagorean triplets.
then m= 6
eg. 3, 4, 5 here 52 = 32 + 42
Thus, m2 – 1 = 36 – 1 = 35
Other Pythagorean triplets are (5, 12, 13), (7, 24, 25),
and m2 + 1 = 36 + 1 = 37
(6, 8, 10), (8, 15, 17) etc.
Therefore, the required triplet is 12, 35, 37.
For any natural number m > 1,
we have (2m)2 + (m2 – 1)2 = (m2 + 1)2. Note : All Pythagorean triplets may not be

So, 2m, m2 – 1 and m2 + 1 forms a Pythagorean obtained using this form. For example
triplet. another triplet 5, 12, 13 also has 12 as a
member.
  DIVISIBILITY TEST

No. Divisibility Test

2 Unit digit should be 0 or even.

3 The sum of digits of no. should be divisible by 3.

4 The no. formed by last 2 digits of given no. should

be divisible by 4.

5 Unit digit should be 0 or 5.

6 No. should be divisible by 2 & 3 both.

7 No. without ones – 2(ones) = no. should divisible

by 7.

8 The number formed by last 3 digits of given no.

should be divisible by 8.

9 Sum of digits of given no. should be divisible by 9.

11 The difference between sums of the digits at even

& at odd places should be zero or multiple of 11.

13 No. without ones +4 (ones digit) = No. should be

divisible by 13.

25 Last 2 digit of the number should be 00, 25, 50 or

75.

Ex.7 Check 119 and 329 is divisible by 7 or not.

Sol. (i) 11 – 2 (9) = –7, it is divisible by 7

 119 is divisible by 7

(ii) 32 – 2 (9) = 32 – 18 = 14 is divisible by 7

 329 is divisible by 7

Ex.8 Check 611 is divisible by 13 or not.

Sol. 61 + 4 (1) = 61 + 4 = 65

here 65 is divisible by 13

 611 is divisible by 13

EXERCISE
Q.1 Find the other two numbers for each of the Q.6 (a) 7a + 43b + c = 518, where a, b, c are in
numbers given below, making the three the units place and c < a < b.
numbers Pythagorean triplets. (b) a36 + b8 + c = 317, where a is in the
(a) 6 (b) 15 (c) 50 (d) 3 hundred digit, b is the tens digit and c is
the ones digit.

Q.2 Without adding, find the value of the

Q.7 a38 + b3 + 5c = 745
following -
(a) 1 + 3 + 5
Q.8 a96 – 43c + 402 – b2 = 814
(b) 1 + 3 + 5 + 7 + 9 + 11
(c) 1 + 3 + 5 + 7 + 9 Q.9 a62 – 473 + 2b6 – 105 + 43c = 1106
(d) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Q.10 Fill in the blanks.
Q.3 Find the cube roots of the following numbers (a) The square of any natural number n can
by successive subtraction of numbers : be written as the sum of _____ odd
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, numbers.
397, ........ (b) When divided by 3, a perfect square
(a) 125 (b) 343 (c) 1728 (d) 512 leaves a remainder of ______ or _____.
(e) 1331 Q.11 Investigate the patterns.
13 + 2 3
Q.4 Using the method of successive subtraction, 13 + 2 3 + 3 3
examine if the following numbers are perfect
Q.12 Create pattern.
cubes. If not, find the smallest number which
Investigate what is
must be subtracted from the numbers so as to
1×2×3×4+1
make them perfect cubes. Also, find their
2×3×4×5+1
cube roots. 3×4×5×6+1
(a) 70 (b) 221 (c) 735 (d) 1011 Using this find value of a, b, c, d if
(e) 349 a×b×c×d+1=1681

Q.5 Solve and find values of a, b, c Q.13 Find the values of unknowns.
(a) 2 5 x 4 (b) 4 p q
(a) 4a + 3(6 – 2) + 25 ÷ 5 = 21
+y 5 2 8 + 7 6 8
(b) (15 ÷ 5) + 3 × 4 – b = 17
12 1 0 2 1 r 2 0
(c) a(18 + 3) + 4 × 5 ÷ 2 – 7 = 45 b 4 a 2 a 4 2
(c) (d)
(d) 2 × 3 + 14 ÷ 7 + 6 – 7c = 35 –6 8 5 × 2 a
9 4 3 2  2 c 8 1 b 8 b 2
(e) 48 ÷ 12 ×  of  of  a  = 6
8 3 4 3  5 2 8 4 0
a 8 a 9 2
(f) 10 – [9 – {8 – (7 – 6)}] – c = 3

(Ques. Q.6 to Q.9) Find a, b, c in the following.

 ANSWER KEY
EXERCISE
1. (a) 8, 10 (b) 8, 17 (c) 40, 30 (d) 4, 5 2. (a) 9 (b) 36 (c) 25 (d) 81
3. (a) 5 (b) 7 (c) 12 (d) 8 (e) 11 4. (a) 6, 4 (b) 5, 6 (c) 6, 9 (d) 11, 10 (e) 6, 7
7
5. (a) 1 (b) –2 (c) 2 (d) – 3 (e) (f) 5 6. (a) a = 5, b = 9, c = 4 or a = 6, b = 8, c = 4 (b) a = 2, b = 7, c = 3
3
7. a = 6, b = 5, c = 4 8. a = 8, b = 6, c = 2 9. a = 9, b = 8, c = 6 10. (a) n (b) 0, 1
3 3 3 3 3 3 2
11.1 + 2 = 9 = 3 ; 1 + 2 + 3 = 36 = 6 12. a = 5, b = 6, c = 7, d = 8
13. (a) x = 7, y = 9 (b) q = 2, p = 5, r = 2 (c) a = 3, c = 5, b = 8 (d) a= 6, b = 5