write the smallest possible value of x if 800x6 is divisible by 11.
Answers
Answer:
The number 11 divides only those numbers
whose sum of digits occupying odd positions is either equal to the sum of the digits occupying even positions or differ from it by a number which is divisible by 11.
The given number is 74X5.
Sum of digits occupying odd positions = 7+X
Sum of digits occupying even positions = 4+5 = 9
For divisibility of 74X5 by 11, 7+X should equal 9. That is, 7+X = 9. This gives X = 2
Applying the other condition for divisibility by 11, we have to take the difference between 7+X and 9. The difference is 7+X-9 = X-2. For X -2 to be divisible by 11, the lowest value of X is 13. But this solution is not admissible because 13 is a 2-digit number and X is to be a single digit number.
Hence, the possible value of X is 2 (Proved)
Proof 2:
Here we divide 74X5 by 11. On first division, quotient= 6 and remainder is 8. Next step is try for X all the single digits from 9 to 1. Taking 9 as the divisor, the dividend is 89. On division by 11, we get 1 as remainder. Bringing down 5, we have 15 and 15 is not divisible by 11. So X = 9 does not make 7495 divisible by 11. Next you try with 8 and then 7 and the remaining digits. When X=2, we have 82 which after division by 11 gives 5 as remainder. 55 is divisible by 11. So the value of X =2 makes 7425 divisible by 11. No need to try with 1.
Hence the possible value of X = 2
Step-by-step explanation:
800x6= 4800
4800 divide 11
no