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Question
A mason has 18 red bricks and 14 orange bricks. In how many different ways can he stack these bricks such that each stack has the same number of bricks and no stack has less than 2 bricks?
A mason has 18 red bricks and 14 orange bricks. In how many different ways can he stack these bricks such that each stack has the same number of bricks and no stack has less than 2 bricks?
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Solution
The correct option is A 1
The mason wants to arrange 18 red bricks and 14 orange bricks in equal stacks.
This is possible if and only if the size of each stack is a common factor of 18 and 14.
Thus, the number of ways this can be done = the number of common factors of 18 and 14.
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 14 = 1, 2, 7, 14
Common factors of 18 and 14 = 1 and 2
But it is given in the question that no stack should have less than 2 bricks.
So the storing can be done in only 1 way by keeping two bricks per stack.
The mason wants to arrange 18 red bricks and 14 orange bricks in equal stacks.
This is possible if and only if the size of each stack is a common factor of 18 and 14.
Thus, the number of ways this can be done = the number of common factors of 18 and 14.
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 14 = 1, 2, 7, 14
Common factors of 18 and 14 = 1 and 2
But it is given in the question that no stack should have less than 2 bricks.
So the storing can be done in only 1 way by keeping two bricks per stack.
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