Three cubes has integer edge lengths. If the sum of their surface area

Let sides of cubes are a, b and c

Surface Area Sum \(= 6a^2+6b^2+6c^2 = 6(a^2+b^2+c^2) = 564\)

i.e.

\((a^2+b^2+c^2) = 94\)

i.e. Sum of three squares = 94

Since \(2^2+3^2+9^2 = 94\)

So Sum of the volumes = \(Side^3 = (a^3+b^3+c^3) = (2^3+3^3+9^3) = 764\)

i.e. Option C, D and E are Eliminated

Since \(3^2+6^2+7^2 = 94\)

So Sum of the volumes = \(Side^3 = (a^3+b^3+c^3) = (3^3+6^3+7^3) = 586\)

i.e. Option B Eliminated

Answer: Option A
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Hello sir,

Just a small question?
How can we find the solution of sum of square of 3 numbers = 94?

I got the first one by hit and trial but how to get second solution?

Thanks
Lakshay

Lucky1994

From the options, we can conclude that only 2 such triplets are possible.

Suppose a≤b≤c

\(a^2+b^+c^2= 94\)

\(\frac{a^2+b^2+c^2}{3} = 31.33\)

hence \(36≤c^2<92\)

c can take only 4 values 6,7,8 or 9

Case 1- \(c^2=81\)

We get,\( a^2+b^2=13\)

since \(\frac{a^2+b^2}{2}\) is 6.5, \( b^2\) can take only 1 value(9).

\(a^2=13-9=4\) (all are perfect squares)

Case 2 - \(c^2=64\)

We get \(\)a^2+b^2=30

\(b^2\)(>15) can take 2 values 16 or 25

1.\( b^2=16; a^2=14\) (rejected)
2. \(b^2=25; a^2=5\)(rejected)

Case3- \(c^2=49\)

We get \(a^2+b^2 = 45\)

b^2(>22.5) can be 25 or 36

1. \(b^2 =25; a^2=20\) (not a perfect square)
2. \(b^2=36'; a^2=9\)(All are perfect squares)\

We don't need to check last case, as we got 2 triplets .

Lucky1994
HI Lakshay

SInce all options suggest that there are two such possible triplets, I started checking if the biggest square subtracted from 94 leaves a number which can further be broken into two perfect squares and got success in such trials.

There are always mathematical ways but in Test situation, it's unlikely to calmly do maths therefore it's important to learn to play with the values. I just played with values and when I subtracted 81 from 94 then got 13...

13 = odd = even+odd = 4+9

Likewise checked for other perfect square 94-64 = 30 but it didn't work as I couldn;t think of two square while I had to think only from 1^2 to 5^2

Then checked 94-49 = 45 which worked when I subtracted 9 from it to leave 36
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\(a^2 + b^2 + c^2 = 94\)
we can have, \(a=3, b=6, c=7\)....(i)
or we can have \(a=2, b=3, c=9\)......(ii)

For case (i), volume = \(3^3 + 6^3 + 7^3 = 586\)
For case (ii) volume = \(2^3 + 3^3 + 9^3 = 764\)

Answer: A

We can let a, b and c be the edge lengths of the three cubes, and we need to determine the possible values of a^3 + b^3 + c^3. We can create the equation for the combined surface area of the cubes as:

6a^2 + 6b^2 + 6c^2 = 564

a^2 + b^2 + c^2 = 94

Since the sum of the 3 squares is less than 100, each edge must be less than 10. Now, let’s let a > b > c. If a = 9 (the largest integer it can be), we have:

81 + b^2 + c^2 = 94

b^2 + c^2 = 13

At this point, we see that b could be 3 and c could be 2. If that is the case, then a^3 + b^3 + c^2 = 9^3 + 3^3 + 2^3 = 729 + 27 + 8 = 764 cm^3 (notice that we can eliminate choices C, D, and E).

Since we still need to find the other possible value of a^3 + b^3 + c^3, let’s let a = 8 and we have:

64 + b^2 + c^2 = 94

b^2 + c^2 = 30

However, there are no integer solutions for b and c.

If b = 7, we have:

49 + b^2 + c^2 = 94

b^2 + c^2 = 45

At this point, we see that b could be 6 and c could be 3. If that is the case, then a^3 + b^3 + c^2 = 7^3 + 6^3 + 3^3 = 343 + 216 + 27 = 586cm^3. We see that the correct answer is A.

Answer: A

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6a^2 + 6b^2 + 6c^2 = 564

a^2 + b^2 + c^2 = 94

The different possibilities being a=9 , b= 3 , c=2
other possibility being a=6 , b=7 , c=3
Which gives us the possibilities 764 and 586
Therefore IMO A