For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form a square, and each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it. If the bottom of the layer has 81 boxes and the top of the layer has only 1 box, how many boxes are in display? A. 236 B. 260 C. 269 D. 276 E. 285

I wouldn't say that. I rather appreciate how GMAC achieves this kind of stuff. What is the fun in asking "What is the sum of squares of integers from 1 to 9?". That's the beauty of this exam. You need to fall in love with this to be able to achieve your elite score. Bunuel can you add the tag as GMATPrep please? Thank you!

I did not know that this one is from GMAT Prep. Added the tag. Thank you.

I thought so. I got this one in the Exam 4 yesterday. Though I did not get it correct

We see that the bottommost layer has 9 x 9 = 81 boxes and the topmost layer is 1 x 1 = 1 box. Therefore, each layer must be a perfect square number of boxes from 1 to 81, inclusive. That is, if we count the layers from top to bottom, they have 1, 4, 9, 16, 25, 36, 49, 64, and 81 boxes. Therefore, the total number of boxes is: 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 Instead of adding the actual numbers, let’s add their units digits: 1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 = 5 + 15 + 5 + 15 + 5 = 45 We see the sum must have a units digit of 5; therefore, choice E is the correct answer. Answer: E

As per the given scenario, No. of boxes in the bottom-most layer = 81 = 9^2 For the layer just above the bottom-most layer, No. of boxes = (9-1)*(9-1)=8^2 Similarly, for the next layer, No. of boxes = 7^2 . . . . No. of boxes in the Top-most layer = 1^2 Total No. of boxes = 1^2 +2^2 + 3^2 + 4^2 + ...............+9^2 = 285 Hence E

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