1.1.1 Introduction
Numbers have played important role in the development of the mankind. This still forms a perfect play ground for activities of children.
1.1.2 Numbers in general form – Playing with Numbers
Consider the number 45. We write this as
45 = 40 + 5 = (4 x 10) + (5 x 1)
Similarly,
34 = 30 + 4 = (3 x 10) + (4 x 1)
In the way,
354 = 300 + 50 + 4 = (3 x 100) + (5 x 10) + (4 x1)
Now let us consider a 9 digit number, 123456789. we can write it as,
123456789 = 100000000 + 20000000 + 3000000 + 400000 + 50000 + 6000 + 700 + 80 + 9
= (1 x 100000000) + (2 x 10000000) + (3 x 1000000) + (4 x 100000) + (5 x 10000) + (6 x 1000) + (7 x 100) + (8 x 10) + (9 x 1)
= (1 x 10^{8}) + (2 x 10^{7}) + (3 x 10^{6}) + (4 x 10^{5}) + (5 x 10^{4}) + (6 x 10^{3}) + (7 x 10^{2}) + (8 x 10^{1}) + (9 x 10^{0})
This is called the base 10 representation of the given numbers or the generalised form of the number.
Consider, for example 136. We can write this in the generalised form as:
136 = (1 x100) + (3 x 10) + (6 x 1)
We see that, 6 is associated with 1; 3 is associated with 10; and 1 is associated with 100. This is the reason, 6 is called the digit in the unit’s place; 3 is the digit in the ten’s place and 1 is the digit in the hundred’s place.
Playing with Numbers – Exercise 1.1.2 – Class VII
 Write the following in the generalised form:
i. 39
Solution:
39 = (3 x 10) + (9 x 1) = (3 x 10^{1}) + (9 x 10^{0})
ii. 52
Solution:
52 = (5 x 10) + (2 x 1) = (5 x 10^{1}) + (2 x 10^{0})
iii. 106
Solution:
106 = (1 x 100) + (0 x 10) + (6 x 1)
= (1 x 10^{2}) + (0 x 10^{1}) + (6 x 10^{0})
iv. 359
Solution:
359 = (3 x 100) + (5 x 10) + (9 x 1)
= (3 x 10^{2}) + (5 x 10^{1}) + (9 x 10^{0})
v. 628
Solution:
628 = (6 x 100) + (2 x 10) + (8 x 1)
= (6 x 10^{2}) + (2 x 10^{1}) + (8 x 10^{0})
vi. 3458
Solution:
3458 = (3 x 1000) + (4 x 100) + (5 x 10) + (8 x 1)
= (3 x 10^{3}) + (4 x 10^{2}) + (5 x 10^{1}) + (8 x 10^{0})
vii. 9502
Solution:
9502 = (9 x 1000) + (5 x 100) + (0 x 10) + (2 x 1)
= (9 x 10^{3}) + (5 x 10^{2}) + (0 x 10^{1}) + (2 x 10^{0})
viii. 7000
Solution:
7000 = (7 x 1000) + (0 x 100) + (0 x 10) + (0 x 1)
= (7 x 10^{3}) + (0 x 10^{2}) + (0 x 10^{1}) + (0 x 10^{0})
 Write the following in the normal form:
i. (5 x 10) + (6 x 1)
Solution:
= 50 + 6 = 56
ii. (7 x 100) + (5 x 10) + (8 x 1)
Solution:
= 700 + 50 + 8
=758
iii. (6 x 1000) + (5 x 10) + (8 x 1)
Solution:
= 6000 + 50 + 8
= 6058
iv. (7 x 1000) + (6 x 1)
Solution:
= 7000 + 6
= 7006
v. (1 x 1000) + (1 x 10)
Solution:
= 1000 + 1
= 1001
 Recalling your earlier knowledge, represent 555 in base 5.
Solution:
[Hint: 4210 are the reminders of 555 when divided by 5]
 What is the representation of 1024 in base 2?
Solution:
1024 = (1100111000)_{2}
1.1.6 Divisibility tests – Playing with Numbers
If a number ends with any of the digits 0, 2, 4, 6, or 8, you immediately say the number is divisible by 2. Why? We write any such number a as a = 10k + r, where r is the remainder when divided by 10. Hence r is one of the numbers 0, 2, 4, 6, 8. We know see that 10 is divisible by 2 and r is also divisible by 2. We conclude that 2 divides a.

Divisibility by 4 [Playing with Numbers]
If a number is divisible by 4, it has to be divisible by 2. Hence the digit in the units place must be one of 0, 2, 4, 6, and 8. But look at the following numbers: 10, 22, 34, 46, and 58. We see that last digit in each of these numbers is as required, yet none of them is divisible by 4. Thus, we can conclude that it is not possible to decide the divisibility on just reading the last digit. Perhaps, the last two digits may help.
If a number has two digits, we may decide the divisibility by actually dividing it by 4. All we need is to remember the multiplication table for 4. Suppose the given number is large, say it has more than 2 digits. Consider the numbers, for example, 112 and 122. We see that 112 is divisible by 4. But 112 = 100 + 22; here 100 is divisible by 4 but 22 is not. Hence 122 is not divisible by 4.
We invoke the following fundamental principle on divisibility:
If a and b are integers which are divisible by an integer m ≠ 0, then m divides a+b, ab and ab.
Now, let us see how does this help us to decide the divisibility of a large number by 4. Suppose we have number a with more than 2 digits. Divide this number by 100 to get a quotient q and remainder r; a = 100q + r, where 0 ≤ r < 100. Since 4 divides 100, you will immediately see that a is divisible by a 4 if and only if r is divisible by 4. But r is the number formed by the last two digits of a. Thus we may arrive at the following test:
STATEMENT 2: A number (having more than 2 digit ) is divisible by 4 if and only if the 2 digit number formed by the last two digits of a is divisible by 4.
Example 4: Check whether 12456 is divisible by 4.
Solution:
Here, the number formed by the last two digits is 56. This is divisible by 4 and hence so is 12456.
Example 5: Is the number 12345678 divisible by 4?
Solution:
The number formed by the last 2 digits is 78, which is not divisible by 4. Hence the given number is not divisible by 4.

Divisibility by 3 and 9 [Playing with Numbers]
Consider the numbers 2, 23, 234, 2345, 23456, 234567. We observe that among these 6 numbers, only 234 and 234567 are divisible by 3. Here, we cannot think of the number formed by the last 2 digits or for those matter even three digits. Note that 3 divides 234, but it does not divide 34. Similarly, 3 divides 456 but it does not divide 23456.
STATEMENT 3: An integer a is divisible by 3 if and only if the sum of digits of a is divisible by 3. An integer b is divisible by 9 if and only if the sum of digits of b is divisible by 9.
Example 6: Check whether the number 12345321 is divisible by 3. Is it divisible by 9?
Solution:
The sum of digits is 1+2+3+4+5+3+2+1 = 21. Hence the number is divisible by 3, but not by 9. In fact 12345321 = (9 x 1371702) + 3.
Example 7: Is 444445 divisible by 3?
Solution:
The sum of digits is 25, which is not divisible by 3. Hence 444445 is not divisible by 3. Here the remainder is 1.

Divisibility by 5 and 10 [Playing with Numbers]
Statement 4: An integer a is divisible by 5 if and only if it ends with 5 or 0. A number is divisible by 10 if and only if it ends with 0.
Example 8: How many numbers from 101 to 200 are divisible by 5?
Solution:
Write the numbers from 101 to 200 which end with 5 and 0. i.e.,
105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200. There are 20 such numbers.
Example 9: Is the number 12345 is divisible by 15?
Solution:
Note that 15 = 3 x 5. Here again the given number 12345 must be divisible by both 3 and 5.
The sum of the digits is 1+2+3+4+5 = 15, which is divisible by 3 and the given number 12345 ends with 5, so it’s also divisible by 5. Therefore, 12345 is divisible by 15.
Example 10: How many numbers from 201 to 250 are divisible by 5, but not by 3?
Solution:
The numbers which are divisible by 5 from 201 to 250 are 205, 210, 215, 220, 225, 230, 235, 240, 245, 250. Now let us compute the digital sum of these numbers:
205 ——2+0+5 = 7, not divisible by 3
210 ——– 2+1 = 3, divisible by 3
215 ———2+1+5 = 8, not divisible by 3
220———–2+2+0 = 4, not divisible by 3
225———2+2+5 = 9, divisible by 3
230———2+3+0 = 5, not divisible by 3
235———2+3+5 = 10, not divisible by 3
240———2+4+0 = 6, divisible by 3
245———2+4+5 = 11, not divisible by 3
250———–2+5+0 = 7, divisible by 3

Divisible by 11 [Playing with Numbers]
STATEMENT 5:
Given number n in decimal form, put alternatively – and + signs between the digits and compute them sum. The number is divisible by 11 if and only if this sum is divisible by 11. Thus a number is divisible by 11 and only if the difference between the sum of the digits in odd places and the sum of digits in even places is divisible by 11.
Example 11: Is the number 23456 divisible by 11?
Solution:
Observe that 23+45+6 = 4 and hence not divisible by 11. The test indicates that 23456 is not divisible by 11.
A palindrome is a number which leads the same from left to right or right to left. Thus a palindrome is a number n such that by reversing the digits of n, you get back n. For example, 232 is a 3 digit palindrome; 5445 is a 4 digit palindrome.
Example 12: Find all 3 digit palindromes which are divisible by 11.
Solution:
A 3 digit palindrome must be of the form aba, where a≠0 and b are digits. This divisible by 11 if and only if ab+a = 2ab is divisible by 11.
This is possible only if 2ab = 0, 11 or 11. Since a≤1 and b≤9, we see that 2ab ≥ 2(1)(9) = 29 = 7 > 11. Hence, 2ab = 11 is not possible. Suppose 2ab = 0. Then 2a = b. Thus, a = 1,b = 2 ; a = 2, b = 4; a=3, b = 6; and a = 4, b = 8 are possible.
We get the numbers, 121, 242, 363, 484.
For a = 6, b = 1, we see that 2ab = 121 = 11 and hence divisible by for which 2ab is divisible by 11. We get four more numbers 616, 737, 858 and 979.
Thus required numbers are 121, 242, 363, 484, 616, 737, 858 979.
Example 13: Prove that 12456 is divisible by 36 without actually dividing it.
Solution:
First notice that 36 = 4 x 9. So it is enough if we prove 12456 is divisible by 4 and 9 both. The last 2 didgits of the given number 12456 is 56 which is divisible by 4 and it is left to show that, the given number 12456 is divisible by 9. Let us find the sum of digits of the given number 12456, i.e., 1+2+4+5+6 = 18, which is divisible by 9. Thus, the number 12456 is divisible by 36.
Playing with Numbers – Exercise 1.1.6 – Class VIII
 How many numbers from 1001 to 2000 are divisible by 4?
Solution:
We know that we have 25 numbers from 1001 to 1100. So we have 25X10 times = 250 numbers from 1001 to 2000 are divisible by 4.
 Suppose a 3digit number abc is divisible by 3. prove that abc+bca+cab is divisible by 9.
Solution:
Let 3 digit number abc be 123, which is divisible by 3. Niw we have to prove, abc+bca+cab i.e., 123+231+312 is divisible by 9.
(1+2+3)+(2+3+1)+(3+1+2) = 6+6+6 = 18. Therefore abc+bca+cab is divisible by 9.
 If 4a³b is divisible by 11, find all possible values of a+b.
Solution:
The number is divisible by 11 if and only if the sum is divisible by 11.
Therefore, the sum of 4a+3b must divisible by 11.
i.e., sum of 7ab = 7(a+b) must be divisible by 11.
Thus, a+b = 18 and a+b = 7.
 Prove that a 4digit palindrome is always divisible by 11?
Solution:
Suppose X has the digits ‘abba’, then X can be expressed as:
X= 1000a + 100b + 10b + a
= 1001a + 110b
= 11(91a + 10b) where (91a+10b) is an integer.
Therefore, X is definitely divisible by 11.
Therefore, a 4 digit palindrome is always divisible by 11.
 Use the digits 4,5,6,7,8 each once, construct a 5 digit number which is divisible by 132.
Solution:
We have 132 = 11 x 4 x 3. Now we have to construct a 5 digit number using 4,5,6,7,8 each once such that it must be divisible by 11, 4 and 3.
Step 1: The last 2 digits of a required 5 digit number must be 56, 76, 84 and 64. So as to get divided by 4.
Step 2: To get divided by 3, the sum of the required numbers must be divided by 3
4+7+8+5+6 = 30
Step 3: To get divided by 11 , the sum of the required numbers must be divisible by 11.
48+57+6 = 0, possible
58+47+6 = 0, possible
67+58+4= 0 possible
57+68+4 = 0 possible
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