A nice problem!

I submitted a problem to brilliant which was unfortunately rejected. I thought it would be nice to share it with all of you. Here it is : 7\small 7-digit numbers (without repetition of digits) are formed using the digits {1,2,3,4,5,6,7} \small \{1,2,3,4,5,6,7 \} . Let S\small S be the set of all such numbers. How many pairs of (a,b)S\small (a,b) \in S exist such that bb is a multiple of aa given that ab \small a \neq b

Note by Vikram Waradpande
6 years, 3 months ago

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This is basically a case checking:

We can have the multiples in the range [2,7] [2,7]

When the multiple is 22 it is not possible, because when you multiply 44[in the number] and 22 , the resulting digit can be either 88 or 99 [The greatest carry over can be only 11 ].

When the multiple is 33, the digit 66 when multiplied by 33 can only give either 8,9,08,9,0 which is again not possible.

When the multiple is 44, the digit 22 in the number when multiplied will produce the corresponding digit 8,9,08,9,0 since highest carry over is 22. This is also not possible.

When it is 55: Now notice that the 1×5=51 \times 5=5, 3×53 \times 5 gives us another digit 55 in the multiplication of the the 77 digit number. Also 7×57 \times 5 gives us the same.Also 5×55 \times 5 gives us another 55 to deal with. Now out of these we need one of them to stay 55 and other to either become 66 or 77. But, the other if it is distinct from other numbers will exceed 77. So no numbers here either.

When it is 66:

1×61 \times 6 and 6×66 \times 6 gives us 66.

So to preserve non-repetition, on one has to be 66 and the other 77.

First note the first digit of aa has to be 11 or our multiple will be out of the domain. This implies that the last digit has to be 66 [carry over concept].

So, if a pair exists in this case, then aa has to have 11 in the first place and 66 in the last.

Now the second last digit that can survive[stay in our range] the carry over 33 derived from the last digit is either 22 or 33. If it is 22 then the number next to 11 can only be 33 since 3×63 \times 6 will gives us a carry over 11. But, wait the other numbers left [4,5,74,5,7 ] will ensure this does not happen.

So, the number before 66 can be 33 and the number after 11 has to be 22.

The digit before 33 cannot be 77 because the numbers to be placed before 77 are only 55 or 44. If it is 55 we have a repetition of the digit 44 and if the digit before 7 is 4, then we have an 88 in the resulting digit. So, 77 must come after 22. Great, this is not possible since that will make our 2nd digit of bb as 66, but 66 is already the last digit.

Now when the multiple is 77 : It is obviously not possible since the first digit will be greater than 88 no matter how you place the digits.

There might be some parts that are incoherent, and I might have easily missed something, in any case let me know if the answer to the question is not 00.

Aditya Parson - 6 years, 3 months ago

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The answer is correct and I like the solution as well!

Vikram Waradpande - 6 years, 3 months ago

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Every number is of the form 9x+1, and since the max we can multiply aa is by 7, there are no such numbers. Nice problem.

Peiyush Jain - 6 years, 3 months ago

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Yeah I had the same solution. All numbers are 11 mod 99. And ba<7 \large \frac{b}{a} \small < 7 . So basically there cannot exist another number that is 11 mod 99

Vikram Waradpande - 6 years, 3 months ago

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Did they give you a reason for rejection?

Peiyush Jain - 6 years, 3 months ago

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@Peiyush Jain They said it did not fit in any of the problem categories.

Vikram Waradpande - 6 years, 3 months ago

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@Vikram Waradpande I see. Thanks.

Peiyush Jain - 6 years, 3 months ago

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