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: $4 x 5 y 790$
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 $| S 1 - S 2 |$ 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 $x + y = 1$ 8, then $18 / 11$ gives a remainder of 7 ( $18 = 11 \times 1 + 7$ ).
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).