With # and & each representing different digits in the problem below,

With subtraction principal, on the units place & - # = 7 and and tens place & - # = 6 this means & is smaller than # so it borrows from the tens digit and subsequently hundred's digit. Now Hundred's digit is # and the result is 6, which is left after the tens digit borrowed 1 from it => # = 7

Thus, at unit place, & - # = 7 => & - 7 = 7 => & was 14 as it borrowed from the tens digit; this means its actual value is 4

Hence, B is the answer

Answer B.
744 - 77 = 667

Did by plugging value.
1) If & = 3, then # becomes 6 but 633 - 66 = 567 not 667. So Not A.
2) If & = 4 then # becomes 7 and 744 - 77 = 667.
Hence B.

Right on SherLocked2018, used the same method, its faster too

Rewrite

667
\(\,\)## +
---------
#&&

# could be either \(6\) or \(7\) - if there is a CARRY-OVER
either way there is a CARRY-OVER, so # is \(7\); & is \(4\)

Answer B

VERITAS PREP OFFICIAL SOLUTION

Now, many would look at this problem and think “I don’t know how to solve problems like that…”, as it’s not a classic “Algebra” problem, but it’s not a straight-up “Subtraction” problem, either. It uses the common GMAT themes of Abstraction and Reverse-Engineering to test you conceptually to see how you think critically to solve problems. And in true Eminem-mocking form, the key to a complicated-looking problem like this is a lot more mainstream than it is advanced. You have to just get started playing with the numbers, testing possibilities for # and & and seeing what you learn from it.

When GMAT students lament that “I don’t know what tools to use” to start on a tough problem, they’re often missing this piece of GMAT wisdom – *that’s* the point. You’re supposed to look at this with some trial-and-error like you would in a business meeting, throwing some ideas out and eliminating those that definitely won’t work so that you can spend some more time on the ones that have a good chance. In this case, throw out a couple ideas for #. Could # be 5? If it were, then you’d have a number in the 500s and you’d subtract something from it. There’s no way to get to 667 if you start smaller than that and only subtract, so even with pretty limited information you can guarantee that # has to be 6 or bigger.

And by the same logic, try a value like 9 for #. That would give you 900-and-something, and the most that ## could be is 99 (the largest two-digit number), which would mean that your answer would still be greater than 800. You need a number for # that allows you to stay in the 667 range, meaning that # has to be 6 or 7. That means that you’re working with:

6&& – 66 = 667

or

7&& – 77 = 667

And just by playing with numbers, you’ve been able to take a very abstract problem and make it quite a bit more concrete. If you examine the first of those options, keep in mind that the biggest that & can be is 9, and that would leave you with:

699 – 66 = 633, demonstrating that even at the biggest possible value of &, if # = 6 you can’t get a big enough result to equal 667. So, again, by playing with numbers to find minimums and maximums, we’ve proven that the problem has to be:

7&& – 77 = 667, and now you can treat it just like an algebra problem, since the only unknown is now 7&&. Adding 77 to both sides, you get 7&& = 744, so the answer is 4.

More important than this problem, however, is the takeaway – GMAT problems are often designed to look abstract and to test math in a different “order” (here you had two unknowns to “start” the problem and were given the “answer”), and the exam does a masterful job of taking common concepts (this was a subtraction problem) and presenting them to look like something you’ve never seen. The most dangerous mindset you can have on the GMAT quant section is “I don’t know how to solve problems like this” or “I’ve never seen this before”, whereas the successful strategy is to take a look at what you’re given and at least try a few possibilities. With symbol problems (like this), sequence problems, numbers-too-large-to-calculate problems, etc., often the biggest key is to go a lot more mainstream than “advanced math” – try a few small numbers to test the relationship in the problem, and use that to narrow the range of possibilities, find a pattern, or learn a little more about the concept in the problem.

If your standard mindset on abstract-looking problems is “I don’t know how to solve problems like that”, both Em and the G-Em-A-T are right to chide you a bit mockingly, as that’s often the entire point of the problem, to reward those who are willing to try (the entrepreneurial, self-starter types) and “punish” those who won’t think beyond the process they’ve memorized. Even if you don’t become a GMAT God, if you follow some of Eminem’s lessons you can at least find yourself saying “Hi, my name is…” over and over again at b-school orientation.

# must be 7 since the difference is given as 667 (hundredths place is 6). Therefore, you have:

7&& - 77 = 667.
7&& = 667 + 77
7&& = 744

Consequently, the answer is (B) 4.

We see that # (in the hundreds digit) represents either 6 (if there is no borrowing from the tens digit) or 7 (if there is borrowing). However, no matter if it’s 6 or 7, it’s also the tens digit of the subtrahend, and since the tens digit of the difference is also 6, there must be borrowing. So # must be 7. In that case, & (in the tens digit) is either 3 or 4. It can be 3 and by borrowing from 7 (in the hundreds place) becomes 13 and 13 - 7 = 6 (if there is no borrow from the units digit). It can be 4 and by borrowing from 7 (in the hundreds place) and by lending 1 to the units digit becomes 13 and 13 - 7 = 6. We can see which one is correct by checking both cases:

733 - 77 = 656 (This is not correct.)

744 - 77 = 667 (This is correct.)

Answer: B

TL;DR

# can only be 6 or 7
If # = 6 => 6&& = 667 + 66 = 7.. (Not possible) => # = 7

7&& = 667 + 77 = 744 => & = 4

Veritas Prep Official Solution

Solution: The big picture: A two digit number is subtracted from a three digit number to give 667. So the three digit number must be a bit larger than 667. This means that the hundreds digit of #&& must be either 6 or 7. It cannot be 8 because you cannot obtain 800+ by adding a two digit number to 667.

Let’s look at both cases:

# is 6: If you subtract 66 from 6&&, you will not get 667 – the largest value you can get is 699 – 66 = 633. So # cannot be 6.

# must be 7.

Now the question is very simple

7&& – 77 = 667

7&& = 667 + 77 = 744

I solved this by trying the answer choices and eliminating whatever won't result in a 667. Luckily the second answer choice was the right one, hence that was quick. For example if we take answer choice 1) 3; 13-7=6, hence # is 6, however 633-66= 567 and not 667, so & can't be 3. if we try second answer choice; &=4, 14-7= 7, and 744-77=667, hence the correct answer is B and we can stop here.

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