International Contest-Game Math Worksheet With Answers - Grade 5 And 6, Kangaroo, 2007 Page 8
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8Bonus 1: Each of the digits from 1 to 9 is used exactly once in the following multiplication
example:
Y
=
7
6
3
2
× × × ×
What digit is denoted by Y?
A) 1
B) 4
C) 5
D) 8
E) 9
Note: Let us denote the digits in the empty positions as follows:
X
Y
Z
A
B
=
7
6
3
2
× × × ×
Solution:
4
2
The digits free to use are 1,4,5,8, 9. The prime factorization of 7632 is
2
⋅
3
⋅
53
. The factor
53 cannot be AB, since 3 is already used in the answer on the right. Hence, XYZ is a three-digit
multiple of 53. Considering the prime factorization, XYZ=53×M, where M is one of the
following numbers: 2, 3, 4, 6, 8, 9, 12, 16, or 18. (M≤18, since 53×18=954 is the greatest three-
digit multiple of 53). Many of these numbers should be rejected because the number XYZ
cannot include any of the digits 7, 6, 3, or 2. On the other hand, if M≥16, then the factors
remaining from the prime factorization have a product that does not exceed 9, thus, cannot be a
two-digit number AB. Taking into consideration all of these restrictions, the only possible
choices for XYZ are
(1) 53×3=159, then AB=48. The multiplication becomes 159×48=7632, and it satisfies the
requirements in the question, so, it is a possible solution.
(2) 53×8=424. Since there is a repetition of the digit 4, it is not a solution.
Finally, the value of the digit Y is 5.
Answer: C
Bonus 2: In a regular die, the faces are numbered by the numbers 1 to 6 and
the sum of the numbers on any two opposite faces is 7. Nick composed a
rectangular prism 2
× × × ×
2
× × × ×
1 using four identical regular dice, with the numbers
on any two touching faces of the dice being equal (see the figure). The
numbers on some faces are shown. Which number must be written on the face
denoted by the (?)?
A) 5
B) 6
C) 2
D) 3
E) not enough
information
Solution:
Using the information that the numbers on any two opposite sides have a sum of 7, it is easy to
see that the top and the bottom faces of the die that contains the “?” cannot be 6 and 1 or 3 and
4, thus, the “?” can only be 5 or 2. But taking into consideration the position of 2 and 3 relative
to each other and to the rest of the digits, as well as the fact that all dice are identical, we obtain
that “?” must be 5.
Answer: A.
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