A hard digit problem
Strike the first digit of the number
, and then add it to the remaining number. Repeat this process until a ten digit number remains. Prove that this number has two equal digits.
I have no idea how to attack this problem. I love a challenge…
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Does it help that a 10-digit number which doesn’t have any 2 equal digits must be a multiple of 9?
Thus your problem can be solved by showing that the final number is not a multiple of 3.
I don’t know— probably :)—, thanks for the tip. I was trying to do stuff with the sum of the digits: if no digits repeat, the sum of the digits of a 10 digit number is 45. So I was trying to show that the actual sum would be something other than 45. Your way seems a lot easier. How did you come up with that fact? I’m going to have to go take a long look at it.
Oh.
Yeah, the divisibility test for 3 (or 9) is that the digital sum is divisible by 3 (or 9).
For example, 21561 is divisible by 3 …. since 2+1+5+6+1 = 15 is divisible by 3.
(Similarly, 21561 is not divisible by 9)
In this case, the digital sum - as you said - is 45, so what you said is perhaps a more general way of looking at it.
I haven’t thought about it any more than that.
So… we know that the digital sum of 7^1996 has some value to begin with.
We will remove the first digit. Add that digit to the remaning whole number.
Assuming there is no carrying involved in the ‘long addition’ we do, then during this process we are adding and taking away the same number to the digital sum - so it remains unchanged.
If we have to carry a digit (or more than one), then in the process of carrying - ‘10′ is becoming ‘1′… so the digital sum is being reduced by 9.
Now, we know that 7^1996 is NOT divisible by 3 (prime factorization)
so the digital sum of 7^1996 will NOT be divisible by 3.
During our calculations to get to the 10-digit number, the digital sum will reduce by some multiple of 9. Therefore, the digital sum will remain NOT a multiple of 3, and thus specifically that the digital sum cannot be 45.
This shows that the digits cannot all be different (in that case the digital sum would be 45, as you said), and there must be two equal digits, as the question requires!
Does this make any sense?
There is a five-digit number. The fifth digit is one fourth of the third digit and one half of the fourth digit. Third digit is one half of the first digit. Second digit is 5 more than the fifth digit. What is that 5-digit number?
It is 86421. Thanks for playing