Page 308 - Start Up Mathematics_8 (Non CCE)

For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 5 if b is either 0 or 5.
(b) A 3-digit number 100a + 10b + c is divisible by 5 if c is either 0 or 5.
For example, 15, 65, 90, 135, 280, etc. are divisible by 5.
V. Test of Divisibility by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 6 if b is 0, 2, 4, 6 or 8 and a + b is divisible by 3.
(b) A 3-digit number 100a + 10b + c is divisible by 6 if c is 0, 2, 4, 6 or 8 and a + b + c is divisible by 3.
For example, 18, 48, 132, 324, 408, etc. are divisible by 6.
VI. Test of Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 9 if a + b is divisible by 9.
(b) A 3-digit number 100a + 10b + c is divisible by 9 if a + b + c is divisible by 9.
For example, 63, 108, 315, 432, 738, etc. are divisible by 9.
VII. Test of Divisibility by 10
A number is divisible by 10 if its digit at the ones place is 0.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 10 if b = 0.
(b) A 3-digit number 100a + 10b + c is divisible by 10 if c = 0.
For example, 40, 90, 270, 480, 950, etc. are divisible by 10.
VIII. Test of Divisibility by 11
A number is divisible by 11 if the difference between the sum of its digits at odd places and sum of its
digits at even places is either 0 or a multiple of 11.
For a number in the generalized form:
A 3-digit number 100a + 10b + c is divisible by 11 if (a + c) – b = 0 or (a + c) – b is a multiple of 11.
For example, 66, 99, 264, 341, 495, etc. are divisible by 11.
Example 4: If 21y5 is a multiple of 9, where y is a digit, then what is the value of y? (NCERT)

Solution: 21y5 is a multiple of 9.
fi 2 + 1 + y + 5 = 8 + y should also be a multiple of 9.
fi 8 + y = 9 or 18 or 27 or 36 and so on.
8 + y = 9 fi y = 1 ; 8 + y = 18 fi y = 10 ; 8 + y = 27 fi y = 19 and so on.
Since y is a digit, so y = 1.

Example 5: If 21y8 is a multiple of 6, where y is a digit, then what is/are the value(s) of y?
Solution: Since 21y8 has an even number ‘8’ in the ones place, it is a multiple of 2. Now, the number
21y8 is a multiple of 6, so it has to be a multiple of 3 also.
21y8 is a multiple of 3.
fi 2 + 1 + y + 8 = 11 + y is a multiple of 3.
fi 11 + y = 12 or 15 or 18 or 21 or 24 and so on.
Note: Multiples of 3, i.e., 3, 6, 9 have not been taken as they give negative values of y.
11 + y = 12 fi y = 1 ; 11 + y = 15 fi y = 15 – 11 = 4
11 + y = 18 fi y = 18 – 11 = 7 ; 11 + y = 21 fi y = 21 – 11 = 10
11 + y = 24 fi y = 24 – 11 = 13 and so on.
Since y is a digit, so y = 1 or 4 or 7.
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