Cayley Problem 2

Week of November 25th, 2021

Cayley (2)

A five digit positive integer is created using each of the odd digits 1, 3, 5, 7, 9 once so that

• the thousands digit is larger than the hundreds digit,
• the thousands digit is larger than the ten thousands digit,
• the tens digit is larger than the hundreds digit, and
• the tens digit is larger than the units digit.

How many such five-digit positive integers are there?

Question:

Conclusions from Conditions:1 and 3: cannot be W or Y9: cannot be X or V or Z, but can be W or Y5: can be Y7: can be Y

Solution (1/3):

Step 3

Rewrite the conditions:W>X,W>V,Y>X, ANDY>Z

Step 2

Rewrite as:VW XYZV-ten thousands W-thousandsX-hundredsY-tensZ-units

Step 1

Possibilty #2:Where V5 X9Z,Same conditions as possibility #1 ∴2 possible integers too

Solution (2/3):

Step 6

Possibility #3:Where V9 X7Z,1,3,5- can be either V or X or Z ∴3 choices-For each choice, there are 2 for Z and 1 for X (3x2x1=6) ∴6 possible integers

Step 5

Possibilty #1:Where V9 X5Z,7- cannot be X or Z since 7>5, ∴V=71 and 3- can be either X or Z (2 possible integers)

Step 4

ANSWER: C)

Therefore, there are 16 five-digit integers.

Adding all possible integers together from each possibility:2+2+6+6= 16

Possibility #4:Where V7 X9Z,Same conditions as possibility #3 ∴6 possible integers too

Solution (3/3):

Step 9

Step 8

Step 7

Waterloo Math Contest Solution:

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