1f 211x01 is divisible by 3, where x is a digit.find the possible values of x.
Answers
Correct option is
A
1,4,7
We know that the given number
18x71
is divisible by 3.
And, if a number is divisible by 3 then sum of digits must be a multiple of 3.
i.e., 1+8+x+7+1= multiple of 3
x+17=0,3,6,9,12,15…
Here x is a digit, where, x can have values between 0 and 9.
x+17=18 which gives x=1.
x+17=21 which gives x=4.
x+17=24 which gives x=7.
Hence, x=1,4,7.
Answer:
x= 1, 4, and 7
Step-by-step explanation:
Divisibility rule: A number is divisible by three when the sum of all of its digits is divisible by 3.
The numbers divisible by 3 is 3, 6, 9, 12, 15, 18,....
given: 211x01
211001 - The sum of its digits (2+1+1+0+0+1) is 5. Thus, it is not.
211101 - The sum of its digits (2+1+1+1+0+1) is 6. Thus, it is divisible by 3.
211201 - The sum of its digits (2+1+1+2+0+1) is 7. Thus, it is not.
211301 - The sum of its digits (2+1+1+3+0+1) is 8. Thus, it is not.
211401 - The sum of its digits (2+1+1+4+0+1) is 9. Thus, it is divisible by 3.
211501 - The sum of its digits (2+1+1+5+0+1) is 10. Thus, it is not.
211601 - The sum of its digits (2+1+1+6+0+1) is 11. Thus, it is not.
211701 - The sum of its digits (2+1+1+7+0+1) is 12. Thus, it is divisible by 3.
211801 - The sum of its digits (2+1+1+8+0+1) is 13. Thus, it is not.
211901 - The sum of its digits (2+1+1+9+0+1) is 14. Thus, it is not.