divisibility tests from1 to 15
Answers
Answer:
Divisibility rule for 1
Every integer is divisible by 1, so no rules are needed.
Ex: 3 is divisible by 1, 4573842 is divisible by 1.
Divisibility rule for 2
The last digit must be even number i.e. 0, 2, 4, 6, 8.
Ex : 3456 is divisible by 2 –> last digit i.e. 6 is an even number.
343423 is not divisible by 2 –> last digit 3 is not an even number.
All even numbers are divisible by 2
Divisibility rule for 3
The sum of digits in given number should be divisible by 3. This is easy method to find numbers divisible by 3.
Ex : 3789 is divisible by 3 –> sum 3+7+8+9= 27 is divisible by 3.
43266737 is not divisible by 3 –> sum 4+3+2+6+6+7+3+7 = 38 is not divisible by 3.
Divisible by 4 rule
The number formed by last two digits in given number must be divisible by 4.
Ex : 23746228 is divisible by 4 –> 28 is divisible by 4.
674235642 is not divisibility by 4 –> 42 is not divisibility by 4.
Divisibility rules for 5
Last digit must be 0 or 5.
Ex: 42340 is divisible by 5 –> 0 is last digit.
675564 is not divisibility by 5 –> 4 is last digit.
Divisibility by 6
The number must be divisible by 2 and 3. Because 2 and 3 are prime factors of 6.
Ex : 7563894 is divisible by 6 –> last digit is 4, so divisible by 2, and sum 7+5+6+3+8+9+4 = 42 is divisible by 3.
567423 is not divisible by 6 –> last digit is 3, so not divisible by 2. No need to check for 3.
7 Divisibility rule
Twice the last digit and subtract it from remaining number in given number, result must be divisible by 7. (You can again apply this to check for divisibility test of 7.)
Ex :
1. 343 is divisible 7 –> 34 – (2*3) = 28, 28 is divisible by 7.
2. 345343 is given number. 3 is last digit. Subtract 2*3 from 34534.
34534 – (2*3) = 34528, we cannot tell that this result is divisible by 7. So, we do it again.
3452 – (2*8) = 3436, we will do it again.
343 – (2*6) = 331, we will do it again.
33 – (2*1) = 31, 31 is not divisible by 7.
So, 345343 is not divisible by 7.
Don’t forget to check the generalized rule at the end.
Divisible by 8
The number formed by last three digits in given number must be divisible by 8.
Ex : 234568 is divisible by 8 –> 568 is divisible by 8.
4568742 is not divisible by 8 –> 742 is not divisible by 8.
Divisible by 9
Same as 3. Sum of digits in given number must be divisible by 9.
Ex : 456786 is divisible by 9 –> 4+5+6+7+8+6 = 36 is divisible by 9.
87956 is not divisible by 9 –> 8+7+9+5+6 = 35 is not divisible by 9.
Divisible by 10
Last digit must be 0.
Ex : 456780 is divisible by 10 –> last digit is 0.
78521 is not divisible by 10 –> last digit is 1.
Divisible test of 11
Form the alternating sum of digits. The result must be divisible by 11.
Ex : 416042 is divisible by 11 –> 4-1+6-0+4-2 = 11, 11 is divisible by 11.
8219543574 is divisible by 11 –> 8-2+1-9+5-4+3-5+7-4 = 0 is divisible by 11.
Divisibility rule of 12
The number must be divisible by 3 and 4. Because (3* (2^2)) are prime factors of 12.
Ex : 462157692 is divisible by 12 –> last 2 digits 92, so divisible by 4, and sum 4+6+2+1+5+7+6+9+2 = 42 is divisible by 3.
625859 is not divisible by 6 –> last 2 digits 59, is not divisible by 4. No need to check for 3.
Divisibility test for 13
Multiply last digit with 4 and add it to remaining number in given number, result must be divisible by 13. (You can again apply this to check for divisibility by 13.)
Ex :
1. 4568 is not divisible by 13 –> 456 + (4*8) = 488 –> 48 + (4*8) = 80, 80 is not divisible by 13.
2. 593773622 is given number. 2 is last digit. add 4*2 to 59377362 -> 59377370
5937737+ (4*0) = 5937737, we cannot tell that this result is divisible by 13. So, we do it again.
593773 + (4*7) = 593801, we will do it again.
59380 + (4*1) = 59384, we will do it again.
5938 + (4*4) = 5954, we will do it again.
595 + (4*4) = 611, we will do it again.
61 + (4 *1) = 65 , 65 is divisible by 13.
So, 593773622 is divisible by 13.
Don’t forget to check the generalized rule at the end.
Divisibility rule of 14
The number must be divisible by 2 and 7. Because 2 and 7 are prime factors of 14.
Divisibility rule of 15
The number should be divisible by 3 and 5. Because 3 and 5 are prime factors of 15.
I can't answer all but I will try
for 2 it should be an even number
for 3 the sum of it should be divided by 3
for 4 last two number should be divided by 4
for 5 it should contain 0or5 at last
for 6 it should be divided by2&3
for 8 last 3 number should be divided by 8
for 9 its sum should be divided by 9
for 10 it should contain 0 at last