15

Divisibility by 15

  • Method 1:
    1. If all the digits except the digit at the unit place are added.
    2. The result is multiplied by 10.
    3. The result of step 2 is added to the digit at the unit place.
    4. If result of step 3 is large then the step 1, 2 and 3 is repeated in order.
    5. If the result of step 4 is divided by 15 and the remainder is zero then the number is divisible by 15 else not.

  • Method 2:⇒ If the number is divisible by 5 and 3 both then the number is divisible by 15.

Proof:

Let the number be abcde.

abcde can be written as
a×10000 + b×1000 + c×100 + d×10 + e

which can also be written as
a×(15×666+10) + b×(15×66+10) + c×(15×6+10) + d×10 + e

Every power of 10 i.e. 10x when divided by 15 gives 10 as remainder,x>1 and is a positive integer.
Proof for remainder of 10x when divided by 15

100 = 6×15 + 10

1000 = 10×(100)
=10(15×6+10)
=10×15×6 + 100
remainder = 10

10000 = 10×(1000)
=10(15×6+10 + 100)
=10×15×6×10 + 1000
remainder = 10, as 10 is remainder of 1000.

Remainder when divided by 15 of 100 is 10, similarly of 1000 (other product has factor 15) the process continues as we get 100 by this process each time and 100 has remainder 10.

Every product is divisible except a×10 + b×10 + c×10 + d×10 + e

So the number is divisible if the sum is divisible by 15.

The number is divisible by 15 if [(sum of all digits except the last digit)×10 + digit at unit place ] is divisible by 15.
Remainder

  • The remainder is the remainder obtained in method 1.

Examples

  1. Is 340 divisible by 15.
    1. (3+4)×10 + 0 = 70
    2. 70
    3. Remainder when 70 is divided by 15 = 10
    4. The number is not divisible and the remainder is 10.

  2. Is 3483 divisible by 15.
    1. (3+4+8)×10 + 3 = 153
    2. Repeating process for 153 (not necessary) [(1+5)10+3] = 63
    3. Remainder when 63 is divided by 15 = 3
    4. The number is not divisible and the remainder is 3.


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