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SOURCE:COMPETITION Number of Problems: 21. FOR PRINT ::: (Book)
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
Let the squares be labeled , , , and .
When the polygon is folded, the "right" edge of square becomes adjacent to the "bottom edge" of square , and the "bottom" edge of square becomes adjacent to the "bottom" edge of square .
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares , , and will prevent the polygon from becoming a cube with one face missing.
Squares , , , , , and will allow the polygon to become a cube with one face missing when folded.
Thus the answer is .
Another way to think of it is that a cube missing one face has of its faces. Since the shape has faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. ,, and overlap, while squares to do not. The answer is
Let be the largest integer that is the product of exactly 3 distinct prime numbers , , and , where and are single digits. What is the sum of the digits of ?
Since is a single digit prime number, the set of possible values of is .
Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .
Using these values for and , the set of possible values of is
Out of this set, the prime values are
Therefore the possible values of are:
The largest possible value of is .
So, the sum of the digits of is
What is the probability that an integer in the set is divisible by and not divisible by ?
There are integers in the set.
Since every 2nd integer is divisible by , there are integers divisible by in the set.
To be divisible by both and , a number must be divisible by .
Since every 6th integer is divisible by , there are integers divisible by both and in the set.
So there are integers in this set that are divisible by and not divisible by .
Therefore, the desired probability is
Due to the wording of the question, it may be taken as "Find the probability that an integer in said set is divisible by 2 and not 3 EXISTS". One example would be 2, which is not a multiple of 3, thus the probability is 1.
How many positive cubes divide ?
Therefore, a perfect cube that divides must be in the form where , , , and are nonnegative multiples of that are less than or equal to , , and , respectively.
So:
( posibilities)
( posibility)
So the number of perfect cubes that divide is
If you factor You get
There are 3 ways for the first factor of a cube: , , and . And the second ways are: , and .
Answer :
Suppose that and are positive integers such that . What is the minimum possible value of ?
must be a perfect cube, so each power of a prime in the factorization for must be divisible by . Thus the minimum value of is , which makes . These sum to .