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You don't need a formula for this. Consider this arithmetic series:
7,7,7,7,...,7 = 350. In this case, there are 50 terms (all 7's).

Now, the other term in the probelm, 77, is a multiple of 7 - that is, 11*7, or 11 sevens go into 77.

So, subtract 77 from 350 and you get 273. 273 has (50-11) sevens. Now 273 + 77 = 350. This implies that you have 39 sevens + 1 seventy seven = 40 terms. That is, 39 * 7 + 1 * 77 = 350, and #of terms = 40

I don't see any real shortcut to this. But I would not call this method brute force. Within the range of values given in the answer choices, 40 is the only that is correct. There are others, if you include more 77 terms but this will reduce the number of terms well below 30 terms.

Alternatively, for more difficult problems you could use the sum of arithmetic series formula : Sn = n(t1 + tn)/2. This would be useful if you had a large number of terms and you wanted to know how many of one term will go into the sum. Subsequently, you could deduce the number of terms of the other number.

But I definately see your point in that under pressure, it is sometimes difficult to do these problems without, say, more objective methods such as formulas.

No need to do calculations.
Since the sum is 350. Last digit of the sum is 0. This can be possible only if the total number of 7s and 77s is multiple of 10. I mean if 7 or 77 are summed up 10 ( or multiple of 10 times) times, then you will get the sum with last digit 0.

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