A 3 digit number 4X3 is added to 984 to get a 4 digit number 13Y7. If 13Y7 is divisible by 11, then what is the value of (X+Y) ?
[amp_mcq option1=”15″ option2=”12″ option3=”11″ option4=”10″ correct=”option4″]
Let’s perform the addition column by column:
Units place: 3 + 4 = 7. This matches the unit digit of 13Y7. No carry-over to the tens place.
Tens place: X + 8 = Y (plus carry-over from units, which is 0). So, X + 8 = Y. Since Y is a single digit, X + 8 must be less than 10.
Hundreds place: 4 + 9 (plus carry-over from tens) = 13. The sum is 13Y7, which is a 4-digit number starting with 13. This means the sum in the hundreds place is indeed 13, and there was no carry-over from the tens place to the hundreds place.
So, X + 8 = Y, where Y is a digit between 0 and 9.
The possible values for X (a digit between 0 and 9) that result in a single digit Y are:
If X=0, Y = 0+8 = 8. The sum is 1387.
If X=1, Y = 1+8 = 9. The sum is 1397.
If X=2, Y = 2+8 = 10. Y cannot be 10 as it’s a single digit.
So, either X=0, Y=8 (sum=1387) or X=1, Y=9 (sum=1397).
We are given that 13Y7 is divisible by 11.
Using the divisibility rule for 11: The alternating sum of the digits (starting from the right) must be divisible by 11.
For 13Y7: +7 – Y + 3 – 1 = 9 – Y.
For 13Y7 to be divisible by 11, 9 – Y must be a multiple of 11 (0, 11, -11, etc.).
Since Y is a digit (0-9), 9 – Y can range from 9-0=9 to 9-9=0.
The only multiple of 11 in this range is 0.
So, 9 – Y = 0, which means Y = 9.
Now we use the relation X + 8 = Y. Substitute Y=9:
X + 8 = 9
X = 9 – 8 = 1.
So, X=1 and Y=9.
Let’s verify: 413 + 984 = 1397.
Is 1397 divisible by 11? 7 – 9 + 3 – 1 = 0. Yes, it is.
The value requested is (X + Y).
X + Y = 1 + 9 = 10.