Question 1119990
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If a positive integer is divisible by 5, then the 1s digit must be either 5 or 0.


Consider the 4-digit positive integers with different digits, leading digit not zero, and 5 is the largest digit, where the 1s digit is zero.


There are 5 possibilities for the 1000s digit, the numbers 1 through 5, then for each of those possibilities there are four choices for the 100s digit, that is the numbers 1 through 5 less the digit that was selected for the 1000s digit.  Then there are 3 ways to pick the 10s digit, 1 through 5 less the two that have already been used, and then the 1s digit is zero.  So there are 5 times 4 times 3 different 4 digit numbers in this category, a total of 60 different numbers.


Now consider the 4-digit positive integers with different digits, leading digit not zero, and 5 is the largest digit, where the 1s digit is 5.


There are 4 ways to choose the 1000s digit, the numbers 1 through 4; zero not allowed as the lead digit, and 5 will be used as the 1s digit.  Then there are 4 ways to choose the 100s digit, 0 through 4 less the one selected for the 1000s digit. Then there are 3 ways to choose the 10s digit and the 1s digit is 5.  So 4 times 4 times 3 = 48.


Grand total:  108 different 4-digit numbers, lead digit not zero, 5 the largest digit, and divisible by 5.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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