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Algebra Level 4

Find the sum of four positive numbers such that:

  1. The sum of the first, third, and the fourth number exceeds the second by 8.

  2. The sum of squares of the first and the second exceeds the sum of squares of the third and fourth by 36.

  3. The sum of the products of the first two numbers and the last two numbers is 42.

  4. The cube of the first number equals to the sum of the cubes of the other three numbers.


The answer is 18.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Best approach is to go to the last condition, where we have , for numbers a, b, c, d, a 3 = b 3 + c 3 + d 3 , . . . . t h a t i s 6 3 = 5 3 + 4 3 + 3 3 = 216. S o a = 6. a^3=b^3+c^3+d^3,.... that\ is\ \ \ 6^3=5^3+4^3+3^3=216.\ \ \ \ So\ a=6.\\
Next see what is the order from first condition. a+c+d=b+8.
Sum + 8= 18+8=26. So 26/2=13=?+8. So ?=5. Makes b=5.
Second and third condition are satisfied with c and d in any of the last two space.
a + b + c + d = 18. a+b+c+d=\huge\ \ \ \color{#D61F06}{18}.


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