Categories
AMC 8 Math Olympiad USA Math Olympiad

Numbers on cube | AMC-10A, 2007 | Problem 11

Try this beautiful problem from AMC 10A, 2007 based on Numbers on cube. You may use sequential hints to solve the problem.

Try this beautiful problem from AMC 10A, 2007 based on Numbers on cube.

Numbers on cube – AMC-10A, 2007- Problem 11


The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

  • \(16\)
  • \(18\)
  • \(20\)

Key Concepts


Number system

adition

Cube

Check the Answer


Answer: \(18\)

AMC-10A (2007) Problem 11

Pre College Mathematics

Try with Hints


Given condition is “The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same”.so we may say that if we think there is a number on the vertex then it will be counted in different faces also.

can you finish the problem……..

Therefore we have to count the numbers \(3\) times so the total sum will be \(3(1+2+….+8)\)=\(108\)

can you finish the problem……..

Now there are \(6\) faces in a Cube…..so the common sum will be \(\frac{108}{6}\)=\(18\)

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *