Interesting Number Theory Problem
The number x = 1 Number of 0’s=2001 0 0 0 0 0 0 0 0 0 … 0 1 is divisible by which of the following numbers?
- 1 Number of 0’s=10 0 0 0 0 0 0 0 0 0 … 0 1
- 1 Number of 0’s=25 0 0 0 0 0 0 0 0 0 … 0 1
- 1 0 1
- 1 Number of 0’s=1000 0 0 0 0 0 0 0 0 0 … 0 1
- 1 1
- 1 Number of 0’s=12 0 0 0 0 0 0 0 0 0 … 0 1
- 1 Number of 0’s=13 0 0 0 0 0 0 0 0 0 … 0 1
- 1 Number of 0’s=70 0 0 0 0 0 0 0 0 0 … 0 1
- 1 Number of 0’s=22 0 0 0 0 0 0 0 0 0 … 0 1
- 1 0 0 1
Submit the sum of the numbers which divide x as your answer. For example, if you think that the last two numbers divide x , submit 9 + 1 0 = 1 9 .
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2 solutions
Sorry, I think there is a mistake. 1 and 3 are identical and both currently are not divisible by your number.
I am quite sorry
Must have copy pasted something wrong. Forgive me.
the problem is wrong it seems
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not anymore - it was for about 1 hour and then I fixed it. Now about 1/2 the people are getting it correct, and I have checked all of the choices to make sure.
Isn't it unfair that unless you know that obscure rule you can't solve the problem?
It’s not really obscure. It’s like saying that knowing x squared - y squared is the product of sum and differences. It’s something you get through practice and learning concepts. It’s also the generalization of factoring x plus minus y cubed.
Number with n zeroes (A) divisible by number with k zeroes (B) if and only if n-k divisible by 2k+2.
If we subtract (10^(n-k)+1)*B from A we get negative same structured number with n-2k-2 zeroes.
I think you mean divide?
All of these can be expressed as 10 k + 1. x can be expressed as 10 2 0 0 2 + 1 ==>> NOT 10 2 0 0 1 +1 or 10 2 0 0 0 +1. Notice the factoring rule that x a +y a is divisible by any x q + y q where a/q is odd. Therefore, you want to find all EVEN factored exponents of 10 of 2002 in the list, as that means that a/q will be odd. Looking at the list, you see that number 2, 3, and 7 are the only ones which have an even exponent of 10 divisible by 2002. The answer is 12!