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

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?

- \(16\)
- \(18\)
- \(20\)

Number system

adition

Cube

But try the problem first...

Answer: \(18\)

Source

Suggested Reading

AMC-10A (2007) Problem 11

Pre College Mathematics

First hint

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........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/roots-of-equations-prmo-2016-problem-8/
- https://www.youtube.com/watch?v=M_HvBNmPcfU

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