CAT 2025 Slot 3 QA Question 8

First, $10^n$ where n is a positive integer is a 1 followed by n zeros. So, $10^{50}$ would be a 1 followed by 50 zeros. Similarly, $10^{25}$ is a 1 followed by 25 zeros. Now, if I add these two numbers together, $10^{50} + 10^{25} , I imagine it would look like a 1 followed by 25 zeros, then another 1, and then 25 more zeros. So, the number would be something like 100...00100...00, right?

But then we subtract 123 from this sum. Subtracting 123 is going to affect the last few digits of the number. Consider this. The number $10^{50} + 10^{25}$ has a 1 at the 51st digit and another 1 at the 26th digit. All the digits in between are zeros. When we subtract 123, we'll need to borrow from the higher digits because the last few digits are zeros.

Let me write this out more concretely. Let's represent $10^{50}$ as:

1 followed by 50 zeros: 100000000000000000000000000000000000000000000000000

And $10^{25}$ as:

1 followed by 25 zeros: 100000000000000000000000000

Adding these together, we get:

100000000000000000000000000100000000000000000000000000

Now, subtracting 123 from this number. Let's think about how subtraction works with such large numbers. The last digit is a zero, so subtracting 3 would require borrowing. Similarly, subtracting 2 in the tens place and 1 in the hundreds place would also require borrowing.

Let me represent the number as a string of digits to make it clearer. The number is:

1 followed by 25 zeros, then a 1, then another 25 zeros.

So, positions 1 to 50: position 51 is 1, position 26 is 1, and the rest are zeros.

When we subtract 123, we need to subtract from the last three digits. But since all the digits from position 1 to 25 are zeros, subtracting 123 will turn the last three digits into something else, and we'll have a chain of borrowing.

Let me try to write this out:

Starting from the end:

Similarly, the tens place is 0, so we have to borrow from the hundreds place, which is also 0, and this borrowing will propagate all the way up to the 26th digit, which is 1.

So, borrowing 1 from the 26th digit (which is 1), it becomes 0, and then each subsequent digit we borrow from becomes 9 because we're effectively subtracting 1 from the higher digit and adding 10 to the current digit.

So, starting from the 26th digit, which is 1, we borrow 1, making it 0. Then, each digit from 25 down to 1 will become 9 because of the borrowing chain.

Wait, let me clarify. The 26th digit is 1, so when we borrow 1 from it, it becomes 0. Then, the 25th digit, which was 0, becomes 10 after borrowing, but since we need to subtract 123, which affects the last three digits, the borrowing only affects the digits from 25 down to 1.

Wait, maybe I'm overcomplicating it. Let's think about the number as:

1 followed by 25 zeros, then 1, then 25 zeros.

So, the number is:

1 000000000000000000000000 1 00000000000000000000000000

Now, subtracting 123:

We need to subtract 123 from the last three digits, which are zeros. So, starting from the units place:

So, the 26th digit becomes 0, and each digit from 25 down to 1 becomes 9 because each borrow turns a 0 into a 9 until we reach the digit we're subtracting from.

So, the last three digits become:

After borrowing, the hundreds place (which was 0) becomes 9, the tens place becomes 9, and the units place becomes 10.

Wait, no. Let me think again.

When subtracting 123, we have:

So, the number becomes:

1 followed by 24 nines, then 0, followed by 25 nines, then 10 - 3 = 7 in the units place.

Wait, no. Let me structure it properly.

Original number: 1 followed by 25 zeros, then 1, then 25 zeros.

After subtracting 123: