Division of Integers

Integers of Class 7

Rule I: The quotient of two integers both positive or both negative is a positive integer equal to the quotient of the corresponding absolute values of the integers.

Rule II: The quotient of a positive and negative integer is a negative integer and its absolute value is equal to the quotient of the corresponding absolute values of the integers.

PROPERTIES OF INTEGERS IN DIVISION

  • If a and b are integers, then a ÷ b is not necessarily an integer.
  • If a is an integer different from 0, we have a ÷ 1 = a.
  • If for every integer a, we have a ÷ 1 = a.
  • If a is a non-zero integer, then 0 ÷ a = 0.
  • If a is an integer, then a ÷ 0 is not defined (it is not ¥)
  • If a, b, c are non-zero integers, then
    a > b ⇒ a ÷ c > b ÷ c, if c is positive.
    a > b ⇒ a ÷ c < b ÷ c, if c is negative.

DIVISIBILITY TEST :

As such there is not defined general rule for checking the divisibility. But for some different divisor we have unique rule of divisibility of numbers. We will discuss the rule of divisors from 2 to 11.

Divisibility by 2

Rule: Any number, the last digit of which is either even or zero (0), is divisible by 2.

e.g . 12, 86, 130, 568926 etc.

Examples of numbers that are do not pass this divisibility test because they are not even or unit

place is not even are

3 , -103 , 1.50 , 21

Divisibility by 3

Rule: If the sum of the digits of a number is divisible by 3, the number is also divisible by 3.

e.g.

123 : 1 + 2 + 3 = 6 is divisible by 3, hence 123 is also divisible by 3.

5673: 5 + 6 + 7 + 3 = 21: therefore divisible by 3.

89612: 8 + 9 + 6 + 1 + 2 = 26: 2 + 6 = 8 is not divisible by 3, therefore the number is not divisible by 3.

Examples of numbers that do not pass this test 14→ 1+4 = 5 and since 5 is not divisible by 3, so 14 is also not.

124 → 1 + 2+ 4 = 7 again sum is not divisible by 3, does not work.

100,002,001→ 1 +0 +0 +0 +0 +2 +0 + 0 + 1= 4 so this very large also does not pass this divisibility test.

Divisibility by 4

Rule: IF the last two digits of a number is divisible by 4, then the number is divisible by 4. The number having two or more zeroes at the end is also divisible by 4.

e.g.

562428: 28 is divisible by 4, hence number is divisible by 4.

5300: There are two zeroes at the end, so it is divisible by 4.

134000: As there are more than two zeroes, the number is divisible by 4.

134522: As the last two digit (22) is not divisible by 4, the number is not divisible by 4.

Examples of numbers that are do not pass this divisibility test

113 since the last two digits, 13, are not divisible by 4, the whole number does not pass this divisibility test.

10,941 the last two digits, 41, are not de visible by 4. Therefore, the whole number does not satisfy the rule for 4.

100,002,014 = 14 and 14 is no good, does not work.

-1,011 = 11 so 1,011 fails this test .

The test for 4 makes sense if you just break down the numbers. Think about what this rule says: “All that matters is whether or not the last two digits are divisible by 4.” Let’s look at why this rule is true.

Examine some three digit numbers

124 is the same as 100 + 24, and we know that 100 is divisible by 4 so all that matters here is whether or not 24, or the last two digits, are divisible by 4. The same could be said for any three digit number 224 = 200 + 24, and we know that 200 is divisible by 4 so again all that we’re worried about are these last two digits.

Any multiple of 100 is divisible by four! Whether you’re talking about 300, 700, 1000, 1100, 123,00– All of these multiples of 100 are divisible by 4, which means that all that we ever have to worry about is the last two digits!

Divisible by 5:

Rule: If a number ends in 5 or 0, the number is divisible by 5.

e.g.

1345: As its last digit is 5 is divisible by 5.

1340: As its last digit is 0, it is divisible by 5.

1343: As its last digit is neither 5 nor 0, it is not divisible by 5.

Examples of numbers that fail this divisibility test.

17since the last digit is 7, 17 does not satisfy this rule and is not divisible by 5

118 since the last digit is 8, 118 does not satisfy this rule and is not divisible by 5

-311 → Since the last digit is 1, 311 does not satisfy the rule for 5

Divisible by 6:

Rule: If a number is divisible by 3 and 2, the number is also divisible by 6. If the is composite then divide the number into co-prime form .if the number is divisible by all the prime factors then the number is divisible by the given number.

e.g.

63924: The first condition is fulfilled on the last digit (4) is an even number and also (6 + 3 + 9 + 2 + 4) = 24 is divisible by 3, therefore, the number is divisible by 6.

154: The first condition is fulfilled but not the second, therefore, the number is not divisible by 6.

Examples of numbers that are do not pass this divisibility test

207Fails the test since it’s not even. We don’t even have to see whether the second condition is satisfied since both conditions must be satisfied to pass this test. If only one of the two conditions (divisible by 2 and by 3) are not met, then the number does not satisfy the rule for 6.

241,124 Although this number is even, the sum of its digits are not evenly divided by 6 so this fails the test.

Divisible by 7:

Rule: The last digit of the number is multiplied by 2 and then subtracted from the remaining number; this process is continued till get the smallest number. Then check it whether divisible by 7 or not.

e.g.

112: 11 – 2 x 2 = 7. As 7 is divisible by 7, the number 112 is also divisible by 7.

2961:

Step I: 2961: 296 – 1 x 2 = 296 – 2 = 294.

Step II: 29 – 4 x 2 = 29 – 8 = 21

As 21 is divisible by 7, the number is also divisible by 7.

5527783 8: 5527783 – 8 x 2 ⇒ 5527783 – 16 = 5527767

552776 7: 552776 – 7 x 2 ⇒ 552776 – 14 = 552762

55276 2: 55276 – 2 x 2 ⇒ 55276 – 4 = 55272

5527 2: 5527 – 2 x 2 ⇒ 5527 – 4 = 5523

552 3: 552 – 3 x 2 ⇒ 552 – 6 = 546

54 6: 54 – 6 x 2 = 54 – 12 = 42

As 42 is divisible by 7, the number is also divisible by 7.

Divisible by 8:

Rule: If the lat three digits of a number is divisible by 8, the number is also divisible by 8.

e.g.

1256: As 256 is divisible by 8, the number is also divisible by 8.

135923120: AS 120 is divisible by 8, the number is also divisible by 8.

139287000: As the number has three zeroes at the end, the number is divisible by 8.

Note: the same rule is application to check the divisibility by 125.

Examples of numbers that are do not pass this divisibility test

9,801 since 801 is not divisible by 8, 9,801 is not.

234,516Nope, no good. 516 is not evenly divided by 8 so the whole number fails the test!

-32,344,588 588 does not work, so -32,344,588 does not satisfy the rule for 8!

Divisible by 9:

Rule: if the sum of all the digits of a number is divisible by 9, the number is also divisible by 9.

e.g.

39681: 3 + 9 + 6 + 8 + 1 = 27 is divisible by 9, hence the number is divisible by 9.

456138: 4 + 5 + 6 + 1 + 3 + 8 = 27 is divisible by 9, hence the number is also divisible by 9.

Examples of numbers that are do not pass this divisibility test

292+9 =11. Since 11 is not divisible by 9, 29 is not either.

6,992 Nope, no good. 6+9+9+2 =26 which is not evenly divided by 9 so the whole number fails the test!

Divisibility by 10:

Rule: Any number which ends with zero is divisible by 10 or we know that the number 10 is composite . co-prime factor of this number are 2 and 5.thus we can say if the number is divisible by 5 and 2 it will be divisible by 10.

Divisibility by 11:

Rule: If the sums of digit at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11.

e.g.

3245682: S1 = 3 + 4 + 6 + 2 = 15

S2 = 2 + 5 + 8 = 15

As S1 = S2, the number is divisible by 11.

283712: S1 = 2 + 3 + 1 = 6

S2 = 8 + 7 + 2 = 17

As S1 and S2 differ by 11, the number is divisible by 11.

84927291658: S1 = 8 + 9 + 7 + 9 + 6 + 8 = 47

S2 = 4 + 2 + 2 + 1 + 5 = 14

As (S1 – S2) = 33 is divisible by 11, the number is also divisible by 11.

Examples of numbers that are do not pass this divisibility test

947 (9+7) – 4 = 12 which is not divisible by 11

10,823 (1+8+3) – (0+2) = 12- 2 =10. No, no good. This one fails!

35, 784 = → (3 + 7 + 4) – (5+8) = 14 – 13 = 1. No, does not satisfy the rule for 11!

12,347, 496, 132 = → (1+3+7+9+3) – (2 + 4 +4 + 6 + 3)= 23- 19 = 4

divisibility rules

 

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