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Erased to a multiple of 7 (Posted on 2021-04-21) Difficulty: 2 of 5
Given the number 188188...188 (number 188 is written 101 times). Some digits of this number are crossed out. What is the largest multiple of 7, that could happen?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution | Comment 1 of 5
One way to test divisibility by 7 is to multiply each of the digits from right to left by (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5...) and then take the sum of all the products.  If the resulting checksum is divisible by 7 then the original number is as well.  

For example we can test divisibility of 188188 by 7 as follows:

8(1) + 8(3) + 1(2) + 8(6) + 8(4) + 1(5)

= 8 + 24 + 2 + 48 + 32 + 5 = 119.  

119 is divisible by 7, therefore so is 188188.  

Since there are six digits in 188188 and the cycle of multipliers is 6 long, then adding a sequence of 188188 to the end of the number won't change its divisibility by 7, i.e. since 188188 is divisible by 7, then so is 188188188188 and 188188188188188188, etc.  Any number that has "188" repeated an even number of times. 

So we could simply cross off the first three digits of the number in the puzzle, and be left with a number that is 188 repeated 100 times, which will be divisible by 7. 

To check if we can do better, we can also see that the first three digits of the original number contribute an additional 8(1) + 8(3) + 1(2) = 34 to the checksum, so the original number is 6 mod 7. 

Removing a digit from anywhere in the number shifts everything over by 1, but since 1 and 8 are both 1 mod 7 the only impact is has on the checksum is removing the last digit (of the sum).  So removing a single digit changes the number from 6 mod 7 to 4 mod 7.  Removing a second digit changes it from 4 mod 7 to 1 mod 7.  Only after removing a third digit does it change from 1 mod 7 to 0 mod 7.  So we need to remove three digits. 

But the preceding paragraph has also convinced me that we can simply remove ANY three digits from the number - 8s and 1s contribute the same either way - and what remains will be divisible by 7.  So instead of chopping off the first three digits, we instead remove the first three 1's.  The resulting number is:

888888188188...188 (six 8s followed by 188 repeated 98 times).  


  Posted by tomarken on 2021-04-21 08:33:29
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