Number problem from WXC

fzy

This problem first appeared at WXC, and I solved it using an idea I got from 野菜花's 北京初中数学竞赛题. She suggested me to post it here. So here it is.

 www.ddhw.com

The repeat of a positive integer is obtained by writing it twice in a row (so, for example, the repeat of 254 is 254254). Is there a positive integer whose repeat is a perfect square?

 

Because a computer can solve the problem quicker than us, I would add a little flavor to it.

 

a) What is the general format of such numbers?

b) Without using a computer, what is the smallest such number you can find? (I will post the smallest one I found.)

www.ddhw.com

 

fzy

Good job. There is a minor correction. For (10^m + 1) * k to be a repeat number, k must have m digits. If 10^m + 1 = a^2 * b, a must be >= 7, and therefore the number we want is (10^m + 1) * b * c^2, where 0.1 < c^2 / a^2 < 1. This is what I call a general formula, although it is not general enough because it does not say how to find m. There is probably no general format for m, but it looks like for any odd a which is not a multiple of 3 or 5, m exists. Using the above general formula, the smallest repeat square I found is a repeat of 826446281*16, or 1322314049613223140496. www.ddhw.com

sean9991

The question is equivalent to finding integer m such that 10^m+1 = a^2*b, where a, b are integers with a > 1 and b < 10^m+1.  Then the number we want is (10^m+1)*b = a^2*b^2 = (ab)^2.  www.ddhw.com I don’t think there is any general format for m.  For example, 10^11+1 is divisible by 11^2, 10^21+1 is divisible by 7^2, and 10^39+1 is divisible by 13^2.  www.ddhw.com An interesting case is when m is odd integer (m = 2k+1).  10^m+1 is divisible by 11, and the quotient is 90…9091, where “90” appears k-1 times.  With proper k, 90…9091 can be divisible by 11, i.e., 9k=1 (mod 11).  So k = 5+11n, that is m = 11+22n, n = 0, 1, 2, …   When n = 0, the number we find is (10^11+1)/11^2=826446281. www.ddhw.com

新用户

  谢谢fzy, 如果我们能从别的论坛转载题来，一定会很受欢迎