Honestly, I don't see another way.

With this specific question, you can also assume x=0 and get the correct result, although the question states that this is not allowed. But since this is a dangerous approach, where you have to know that it will lead to the same result as a positive integer, I don't recommend it.

But anyways: if x=0, then 7^3 + 3. 49*7+3=346, so remainder is 1.

I guess it should be 7^{12x+3}+3. The solution explains that the last digits of the powers of 7 repeat in a cyclical manner: 7, 9, 3, 1, 7, 9, 3, 1,...
Meaning, if the exponent is a multiple of 4 + 1 (M4+1 like 1, 5, 9,...), the last digit is 7, for a M4+2 (like 2, 6, 8,...) the last digit is 9, etc.

1 is also a M4 + 1, because 1 = 4\cdot0 + 1.

In this question, it is obvious the answer does not depend on the value of x. But, I am going to tell you a secret: even if they say x is positive, you can still plug in 0 for x, the answer will be the correct one. So, just look at 7^3+3, and find the last digit, which is 6, so the remainder when dividing by 5 is 1.
They just state that x is positive, meaning at least 1, so that you cannot raise 7 to the 12\cdot1+3=15th power.
Since 12x is a multiple of 4, \,\,12x + 3 is a M4+3. In other words, only the remainder is important, you can ignore 12x or just take it as 0.

Answer B.
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PhD in Applied Mathematics
Love GMAT Quant questions and running.

Many good insights above. I'll add: there is actually a whole class of questions on the GMAT that ask either "what is the remainder when x is divided by 5" or "what is the remainder when x is divided by 10"? Both of these questions can be answered just by knowing the units digit of x. The reason why that works should be fairly clear if you do a few division problems (try finding the remainder for 7÷5, 37÷5, and 127÷5).

Once you understand that part of the problem, the question becomes "what is the units digit of (7 ^ 12x+3) + 3?" To figure that out, we first need to know the units digit of (7 ^ 12x+3), since the final +3 will be easy to deal with.

If you have no idea what to do at this point, I think the best thing is just to figure out what the units digit of 7^1 is, then of 7^2, then of 7^3, then of 7^4, etc. It's easy for the first two (7 and 9), but remember that to find the units digit of 7^3, you don't actually need to figure out 49*7. You just need to figure out 9*7. 9*7 is 63 (and by the way, 40*7 is 280 - since that ends in a 0, it won't affect the units digit when we add it to 63 to figure out what 7^3 is). So the units digit of 7^3 is 3. And the units digit of 7^4 is therefore the same as the one of 3*7, or 1. The units digit of 7^5 is 7 again, because 1*7 is 7. Note that once you get to a units digit of 1, that's when the pattern starts repeating. Check out Pansi's chart above.

So assume x=1. The units digit of 7^15 is the same as 7^11, 7^7, and 7^3. That would be 3. Add 3 to that and you get 6. Divide by 5 and you get a remainder of 1.

Final note: 7^x isn't the only base with an interesting units digit pattern:
2^x goes 2-4-8-6
3^x goes 3-9-7-1
4^x goes 4-6
5^x is always 5
6^x is always 6
8^x goes 8-4-2-6
9^x goes 9-1
and by the way, 17^x goes 7-9-3-1 just like 7^x does. So does 27^x. You get the idea. No need to memorize these patterns, just understand how to figure them out.
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