1.A rectangle with integral side lengths has area 23.Find its perimeter.
2.Let n be an integer greater than 1. If the sum of digits of n^2 and n^3 are both a while the sum of digits of n^4 is b, find the smaller possible value of b.
3.What is the remainder when the 2009-digit number 1000...0001 is divided by 11?
Maths
2009-03-16 2:57 am
回答 (1)
2009-03-16 3:23 am
✔ 最佳答案
1.Since 23 is a prime number ,only have one case : 1*23So the perimeter =(1+23)*2= 48
2. the smallest value of b is 1 . Consider 10^2 , 10^3 ,10^4 can get the result.
3. 1000...0001 - 2 = 9999...99( 2008 『9』here)
= 90909...0909 + 2
the remainder is 2 .
2009-03-15 19:25:28 補充：
Correction of Q3:
=11 * 90909...0909 + 2
the remainder is 2.
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