2007-12-09 19:33:45 · 5 answers · asked by lovelystarr_86dl 1 in Science & Mathematics ➔ Mathematics
Using exhaustive search between the numbers 317 and 999, there isn't any that once squared will have the form abcabc.
Not sure how to prove it though.
/m
2007-12-09 19:42:28 · answer #1 · answered by perplexed* 3 · 0⤊ 0⤋
You want N * 1001 to be a square, where 100 <= N <=999
1001 = 990 + 11 = 91 * 11 = 7 * 13 * 11
I'm sorry, but it can't be done. Each of 7, 11, and 13 has to divide N, which means 1001 does, so N can't be as small as you need.
2007-12-09 19:41:46 · answer #2 · answered by Curt Monash 7 · 2⤊ 0⤋
Curt Monash is right.
You need some 3-digit number abc, such that (1001 * abc) is a perfect square. If 1001 had any square factor already, this would be easy (abc = 1001 / sq). But it hasn't, so for any multiple of it to be a square, the multiplier has to repeat each prime factor of 1001. And the smallest multiplier which does that is ... 1001!
2007-12-09 21:01:46 · answer #3 · answered by Anonymous · 0⤊ 0⤋
abcabc
= c + 10b + 100a + 1000c + 10000b + 100000a
= 1001c + 10010b + 100100a
= 1001 (c + 10b + 100a)
= 7 x 11 x 13 x (c + 10b + 100a)
This shows that there is no number of the type abcabc that is a perfect square.
2007-12-09 19:44:26 · answer #4 · answered by Madhukar 7 · 0⤊ 1⤋
calculator. there is one in windows application just put the number xxxx^2 that weird hat means to the power.
2007-12-09 19:36:43 · answer #5 · answered by na 2 · 0⤊ 7⤋