From Mathcounts Mini :
Counting/Paths Along a Grid
From Art of Problem Solving
Counting Paths on a Grid
Math Principles : Paths on a Grid : Two Approaches
Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally?
Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways
Question # 2: How many ways can you move from A to B if you can only move downward and to right?
Solution :
There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B
Friday, May 5, 2023
Friday, March 3, 2023
Sum of All the Possible Arrangements of Some Numbers
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#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)
Questions to ponder: (detailed solutions below)
It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions.
#1:
Camy made a list of every possible distinct four-digit positive integer
that can be formed using each of the digits 1, 2 , 3 and 4 exactly once
in each integer. What is the sum of the integers on Camy's list?
#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)
#3: 2020 Mathcounts state sprint #24
Solutions:
#1:
Solution I:
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.
the four numbers.
2.5 (1000 + 100 + 10 + 1) x 24 = 66660
#2:
Solution I:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976
Solution II:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements) + 96 = 4.5 * 11110 * 24 + 96 = 1199976
Solutions:
#1:
Solution I:
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.
Solution II:
The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrangethe four numbers.
2.5 (1000 + 100 + 10 + 1) x 24 = 66660
#2:
Solution I:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976
Solution II:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements) + 96 = 4.5 * 11110 * 24 + 96 = 1199976
#3: The answer is 101.
Other applicable problems: (answers below)
Answer key:
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\)
#3: 133320
#4: 1333272
#5: 93324
#1: What is the sum of all the four-digit positive integers that
can be written with the digits 1, 2, 3, 4 if each digit must be used
exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)
#2: What is the average (mean) of all 5-digit numbers that
can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly
once? (You can use a calculator for this question.) (2005 AMC-10 B)
#3: What is
the sum of all the four-digit positive integers that can be written with
the digits 2, 4, 6, 8 if each digit must be used exactly once in each
four-digit positive integer?
#4: What is
the sum of all the 5-digit positive odd integers that can be written
with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly
once in each five-digit positive integer?
#5:What is
the sum of all the four-digit positive integers that can be written with
the digits 2, 3, 4, 5 if each digit must be used exactly once in each
four-digit positive integer?
Answer key:
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\)
#3: 133320
#4: 1333272
#5: 93324
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