Advertisements
Advertisements
प्रश्न
31×5 divisible by 3?
Advertisements
उत्तर
31×5 is divisible by 3
⇒ 3 + 1 + x + 5 is a multiple of 3
⇒ 9 + x = 0, 3, 6, 9,
⇒ x = -9, -6, -3, 0, 3, 6, 9,
Since, x is a digit
x = 0, 3, 6 or 9
APPEARS IN
संबंधित प्रश्न
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
If x is a digit such that the number \[\overline{{18x71}}\] is divisible by 3, find possible values of x.
If x is a digit of the number \[\overline {{66784x}}\] such that it is divisible by 9, find possible values of x.
Which of the following statement is true?
If a number is divisible by 9, it must be divisible by 3.
Find which of the following numbers are divisible by 3:
(i) 261
(ii) 777
(iii) 6657
(iv) 2574
If a number is divisible by 6, then it must be divisible by 3
If the sum of digits of a number is divisible by three, then the number is always divisible by ______.
If x + y + z = 6 and z is an odd digit, then the three-digit number xyz is ______.
3134673 is divisible by 3 and ______.
The digit sum of 91 is 10. Therefore, 91 is:
