To the first question, actually, is to buy time, so the regularity view, we accept the suggestion made by Miguel Escobar in the comments, add 1 + 2 + .. + 8 and 2 + 3 + ... + 9 separately, using the "trick" of the matches, taking (1 + 8) * 4 and (2 + 9) * 4, ie 36 and 44. As the scores are multiplied by 10, add 360 + 44 = 404, which is the correct solution (B). For
the second question, obviously the number of divisions must be a common multiple, and the youngest is 30 (D).
The third question is to think a little more, is a type of puzzle sudoku. The 9 can only add 11 with a single number, so it must go in one end and 2 at the intersection.
for 8 Now we have two possible positions. Suppose we place it in the next round, we force the intersection to be 1 and we quickly came to a contradictory situation.
Then the other end is occupied by the 8 and the next intersection by 3.
A quick score leaves us companion of 2 to 5, which requires the 4 at the next intersection, to 7 on the other side and at the intersection and the place that interests us in this setting is 6.
Can there be other configurations (in addition to symmetric)? The answer is no, but in this type of test should not stop to check. When we find a valid configuration, you must respond. And the answer is D (6).
Actually, there is an argument even faster. Once we have placed the 9 and 8 at the ends, the two central circles added 22, the ends 8 and 9 are the sum total to 39, and as the sum of the numbers 1 through 9 is 45, the central should be, as we know, 6.
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