Successive numbers sums
□ □ 6 □ □ □ □ □ □ − 4 □
In a sequence of eleven real numbers, only two are shown above. The product of every 3 successive boxes is 120.
Find the sum of all the numbers in the boxes (including the two already open).
Hint: There are only 3 distinct numbers in the boxes.
The answer is -18.
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2 solutions
Similar solution as @Otto Bretscher
Because every three successive number product is 120(which is not 0) so none of the variables are 0.Also since a b c = b c d so that means a = d and so the varibles repeat every 3 numbers and look what we have done: − 4 □ 6 − 4 □ 6 − 4 □ 6 − 4 □ The missing square is 1 2 0 = 6 x ( − 4 ) ⟹ x = 6 ( − 4 ) 1 2 0 = − 2 4 1 2 0 = − 5 .
And the answer is 3 × 6 + 4 ( − 4 ) + 4 ( − 5 ) = 1 8 − 1 6 − 2 0 = 1 8 − ( 1 6 + 2 0 ) = 1 8 − 3 6 = − 1 8
If four consecutive entries are a , b , c , d , then a b c = b c d = 1 2 0 . Thus a = d and the entries repeat after 3 boxes. Counting back, we see that the first entry is − 4 and the second one is 6 ( − 4 ) 1 2 0 = − 5 . Thus the answer is 3 × 6 + 4 ( − 4 ) + 4 ( − 5 ) = − 1 8 .