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They can all be expressed as themselves to the power one, but to make the problem interesting let's specify that the exponent has to be an integer greater than 2. The highest exponent we are going to need is the highest integer solution to 2p < 1000. Which is in fact p = 9. [29 = 512.] The highest base we are going to need is the highest integer solution to b2 < 1000. This is given by b = 31. [312 = 961.] But the only real way to answer the question is just to get stuck in. 1 = 12
4 = 22
8 = 23
9 = 32
16 = 24 = 42
25 = 52
32 = 25
36 = 62
49 = 72
64 = 26 = 43 = 82
81 = 34 = 92
100 = 102
121 = 112
125 = 53
128 = 27
144 = 122
169 = 132
196 = 142
216 = 63
225 = 152
243 = 35
256 = 28 = 44 = 162
289 = 172
324 = 182
343 = 73
361 = 192
400 = 202
441 = 212
484 = 222
512 = 29 = 83
529 = 232
576 = 242
625 = 54 = 252
676 = 262
729 = 36 = 93 = 272
784 = 282
841 = 292
900 = 302
961 = 312
1000 = 103
Well, they're all there. 41 different numbers in total.
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How many perfect square numbers between 1 to 1000 are?
There are 31 perfect square numbers between 1 and 1000 (including 1).
What is 1 and 2 thousands as a decimal?
Expressed as a decimal, 1 2/1000 is equal to 1.002.
What is the average of all the numbers 1 to 1000?
500.5 Because: 1+2+3+.......998+999+1000 Add 1 to 1000 equaling 1001. This makes a pair. There are 500 more of these so 500 pairs. So 1001x500= 500500 Finally divide it by 1000 because there are 1000 numbers. (1000-1=999+1=1000) Concluding the answer to be 500.5 :)
How many palindromic numbers are there between 1 and 1000?
There are 107 numerical palindromes between the numbers 1 and 1000, starting from 2 to 999.
What are the first 1000 odd numbers?
All the odd numbers between 1 and 2001.
