For a display, identical cubic boxes are stacked in square

Easy calculation:

1) Sum the UNIT digits.
1 + 4 + 9 + 16 + 25 +36+ 49+ 64 +81

2) Agrupate in 10's
1+9
4+6
6+4
9+1
5+ ___

The unit digit must be 5, therefore the correct answer is E.

Thanks bunuel but how did you manage to understand that?
I read the question again and cannot imagine the picture you uploaded in my head at all.
Thanks!

I got confused (and I still am) by the line which says "Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it"

I got that the bottom layer will have 9 x 9 boxes
I also understand that the next level up will have 8 rows of boxes and since this layer also has to form a square hence it needs 8 boxes in the column as well.

What is elusive for me is "and 1 fewer box in each remaining row[/color] than the layer directly below it". Can you please explain again.

Each layer has 1 fewer row and 1 fewer box in each remaining row than the layer directly below it:

1st layer has 9 rows and 9 boxes in each of them. 2nd row has 1 fewer, so 8 rows and each of the remaining 8 rows has 1 fewer box, so 8 boxes in it.

Hope it's clear.

Bunuel,

Do you have more examples like these questions? I am finding it difficult to understand after reading question.

Thanks

~M14

superb explanation Bunnel! i couldnt understand the qs...

Is there any chance you can apply why that 6 is there? I want to make sure I can apply this formula in more complicated cases.

Here's where non-native speakers could have trouble. By display I kept thinking of a computer display and I tried to visualize boxes arranged within the TV, and jumped into the conclusion that this was similar to a problem in the OG (13th Ed. PS 124).
Now, if you don't know the formula for the sum of the first n perfect squares (I actually forgot it on a second attempt), it is just nonsense to sum each square result. plaverbach's approach is the appropriate one. After taking a look at the answers and noticing that only two of them have the same units number, you pray that those are wrong and go ahead and find that unit.

I liked plaverbach approach, as I couldn't understand the question in the first place and choose random wrong answer.
But when I saw the picture posted, I could use the plaverbach approach.

In the above figures ,small circles are akin to cubic boxes and lines are rows. So bottom most layer has 9*9 = 81 boxes . Now this figure has more layers stacked on top of it and each layer has 1 less box(small circle) and 1 less row( line). If you follow this theory then you will notice that no of rows = no of boxes in each row.

So 2nd layer from the top will have 2 rows with 2 boxes each. Similarly top most layer will have 1 row and 1 box.

So the total no of boxes will be : 9* 9 + 8*8 +.....+ 1*1 = 285 .

possibly the worst worded question i've ever read

Hi,
The Q seems to be flawed as 21 cannot be the number of boxes . It is either 81 or 25.

You can imagine this as a huge cube from which steps are made in 2 sides by removing a row of boxes in each layer.
In numerical value,all the layers will be square of integers starting from 1 on top to 5 or 9 in lowermost.

When we add these, it becomes \(1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2\)
Either add all of them or use formula
SUM=\(\frac{n(n+1)(2n+1)}{6}\)=9*(9+1)*(2*9+1)/6=9*10*19/6=15*19=285
E

I had a tough time understanding the question as well

Yeah i thought from each square floor, one block is removed.
solved like-
9^2 -1 + 8^2 -1 + .... + 2^2 -1 + 1 ( since it is said from below floor one block is removed) .
And perplexed over where is the answer.
Now i understood what does it mean. It simply means a square box is missing from row in each subsequent floor.

As we know the series is 9^2+8^2.....+2^2+1
we know there are 5 odd+4 even=odd hence striking off Answer choices A,B,and D now we can sum only the unit digits of the squares
1+4+9+6+5+6+9+4+1=X5 hence answer choice ending with 5 is the Answer

Answer :E

Hi All,

The math behind this question is not difficult - it's just about adding the perfect squares from 1^2 to 9^2, inclusive. While it might take a little time to "visualize" what this question describes, the actual "work" is not too challenging. If a child can do the math, then you should be able to (and faster).

There is an interesting shortcut built into the math though. From the answer choices, you'll notice that almost all of the units digits are unique (6, 0, 9 and 5). This provides a likely shortcut that can help us to avoid a bit of the math.

1
4
9
16
25
36
49
64
81
---
?

The nine numbers above can be "paired up" to create 4 values that end in a 0:
1+ 9 --> ends in 0
4 + 16 --> ends in 0
36 + 64 --> ends in 0
49 + 81 --> ends in 0
and 25

The sum MUST end in a 5 and there's only one answer that matches…

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Or you can just multiply the units of the numbers and add them all together you will get a result of 5. The only answer ending in 5 is the correct one