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Problem 1. One digit of 970405 can be changed to make the result divisible by 225. What is the six-digit number after the modification?

 

Problem 2. For each positive integer n, let S(n) denote the sum of the digits of n. How many three-digit n's are there such that n+S(n)+S(S(n)) is conguent to 0 (mod 9)?

 Oct 7, 2020
 #1
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1. You can change the number to 971504.

 

2. There are 334 3-digit numbers that work.

 Oct 7, 2020
 #2
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I just tested them, and both answers are wrong

 Oct 7, 2020
 #3
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1 - Change 970405 to 970425, because 970425 mod 225 = 0.

 

2 - 108  117  126  135  144  153  162  171  180  189  198  207  216  225  234  243  252  261  270  279  288  297  306  315  324  333  342  351  360  369  378  387  396  405  414  423  432  441  450  459  468  477  486  495  504  513  522  531  540  549  558  567  576  585  594  603  612  621  630  639  648  657  666  675  684  693  702  711  720  729  738  747  756  765  774  783  792  801  810  819  828  837  846  855  864  873  882  891  900  909  918  927  936  945  954  963  972  981  990  999  Total =  100 such numbers.

 

Each of the above numbers = n + S(n) + S(S(n)) mod 9 =0. Example: [189 + S(1+8+9 =18) + S(S(1+8))=9] =216 mod 9 = 0

 Oct 7, 2020
 #4
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100 is also wrong, but thanks for trying

 Oct 9, 2020
 #5
avatar+1372 
+1

Problem 1. One digit of 970405 can be changed to make the result divisible by 225. What is the six-digit number after the modification?

 

970425 / 225 = 4313 laugh

 Oct 16, 2020

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