The correct option is (d) - 7.
[as per provisional answerkey]Solution
To determine the divisibility of a number by 11, we use the divisibility rule: The difference between the sum of digits at odd positions and the sum of digits at even positions must be 0 or a multiple of 11.
Given number: 4x5y790
Positions (from left to right):
1st: 4, 2nd: x, 3rd: 5, 4th: y, 5th: 7, 6th: 9, 7th: 0
Step 1: Calculate the sums of digits at odd and even positions.
Sum of digits at odd positions (S1) = 4+5+7+0=16
Sum of digits at even positions (S2) = x+y+9
Step 2: Apply the divisibility rule.
The difference ∣S1−S2∣ must be 0, 11, 22, etc.
Difference = ∣16−(x+y+9)∣=∣7−(x+y)∣
Step 3: Solve for (x+y).
Case 1: 7−(x+y)=0
x+y=7
Case 2: 7−(x+y)=−11 (since x and y are digits, x+y cannot be large enough to make the difference -22)
x+y=7+11=18
Case 3: 7−(x+y)=11
x+y=−4 (Not possible as x and y are digits)
Step 4: Find the remainder when (x+y) is divided by 11.
If x+y=7, then 7/11 gives a remainder of 7.
If x+y=18, then 18/11 gives a remainder of 7 (18=11×1+7).
In both possible cases, the remainder is 7.
Why the other options are incorrect
- Option (a) - 1: This would require (x+y) to be 1 or 12. If x+y=1, the difference ∣7−1∣=6 (not divisible by 11). If x+y=12, the difference ∣7−12∣=5 (not divisible by 11).
- Option (b) - 3: This would require (x+y) to be 3 or 14. If x+y=3, the difference ∣7−3∣=4. If x+y=14, the difference ∣7−14∣=7. Neither is divisible by 11.
- Option (c) - 5: This would require (x+y) to be 5 or 16. If x+y=5, the difference ∣7−5∣=2. If x+y=16, the difference ∣7−16∣=9. Neither is divisible by 11.
Key Concept
The Divisibility Rule of 11 states that the difference between the sum of digits in odd places and even places must be a multiple of 11 (including 0).