Question 1150794: Find the number of five-digit numbers less than 40000 in which the three three-digit numbers formedIf by consecutive digits are all divisible by 17 or 23. Zero cannot be the first digit in a three-digit number. Thank you.
Answer by greenestamps(13332)   (Show Source):
You can put this solution on YOUR website!
With the given restrictions, I doubt there is a single formal mathematical process for solving this problem.
An Excel spreadsheet and some logical detective work can get you to the answer.
Each consecutive 3 digits in the 5-digit number must be divisible by either 17 or 23, and the first digit can't be 0. So do the following:
(1) Use the fill/series and sort features of excel to make a sorted list of all 3-digit numbers divisible by either 17 or 23. Print out the list for easy reference.
(2) Since the 5-digit numbers are to be less than 40000, the first digit must be 1, 2, or 3. So look at each 3-digit number in the list with first digit less than 4 and see if a 5-digit number can be built with the given requirements.
Here is the start of the task....
102: The first number in the list is 102. We won't be able to make a 5-digit number that meets the requirements using this as the first 3 digits, because the second 3-digit number would have 0 as the first digit.
115: The second 3-digit number in the list is 115. If those are digits 1-3, then digits 2-3 must be "15". There is one 3-digit number in the list with first two digits 15; that number is 153. But there is no number in the list with first two digits 53. So there is no 5-digit number with 115 as the first 3 digits.
119: There is no 3-digit number in the list with first two digits 19.
136: There is one 3-digit number in the list with first two digits 36 -- 368. And there is one 3-digit number in the list with first two digits 68 -- 680.
So we have found our first 5-digit number that meets the requirements of the problem: 13680.
Continue in that fashion starting with the other 3-digit numbers in the list that are less than 400....
My final list (possibly not completely correct) contains 12 numbers that satisfy the conditions.
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