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# Augh Hard

+2
145
4
+475

Using only the digits 7, 8 and 9, how many positive seven-digit integers can be made that are palindromes?

I really don't know how to do this :(

Jul 17, 2020

#1
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7777777 , 7778777 , 7779777 , 7787877 , 7788877 , 7789877 , 7797977 , 7798977 , 7799977 , 7877787 , 7878787 , 7879787 , 7887887 , 7888887 , 7889887 , 7897987 , 7898987 , 7899987 , 7977797 , 7978797 , 7979797 , 7987897 , 7988897 , 7989897 , 7997997 , 7998997 , 7999997 , 8777778 , 8778778 , 8779778 , 8787878 , 8788878 , 8789878 , 8797978 , 8798978 , 8799978 , 8877788 , 8878788 , 8879788 , 8887888 , 8888888 , 8889888 , 8897988 , 8898988 , 8899988 , 8977798 , 8978798 , 8979798 , 8987898 , 8988898 , 8989898 , 8997998 , 8998998 , 8999998 , 9777779 , 9778779 , 9779779 , 9787879 , 9788879 , 9789879 , 9797979 , 9798979 , 9799979 , 9877789 , 9878789 , 9879789 , 9887889 , 9888889 , 9889889 , 9897989 , 9898989 , 9899989 , 9977799 , 9978799 , 9979799 , 9987899 , 9988899 , 9989899 , 9997999 , 9998999 , 9999999 , Total =  81 such numbers

Jul 17, 2020
#2
+475
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Thanks :D

However, is there an easier way than counting everything?

Jul 17, 2020
#3
+1

Since you only have 3 numbers, that means the entire number of palindromes will begin with one of the 3 numbers. And since the 3 numbers can be repeated, then each sequence of palindromes beginning with any of the 3 numbers will have:3^3 palindromes =27. And since you have only 3 numbers, then the total number of palindromes will = 27 x 3 = 81 palindromes. So, if you count the number of palindromes beginning with 7, 8 and 9, you should get 27 for each of the 3 numbers for a total of 81 such numbers.

Jul 17, 2020
#4
+475
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Thank you very much!

Jul 17, 2020