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 September 15th, 2009, 05:40 AM #1 Newbie   Joined: Aug 2009 Posts: 12 Thanks: 0 Algebra... plis help me to solve this questions...... 1. Find the remainder when 7^2005 divide by 5..... 2. The sum of a whole number , N and 2005 can be divided respectively by 11 and 9..Find the lowest value of N. plis help me to solve this questions....
 September 15th, 2009, 07:37 AM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Algebra... 1. Try working out the first few of these until you spot a pattern of repetition. You need only worry about the last digit (why?) for the multiplications. Then figure out where 2005 falls in that pattern.
 September 15th, 2009, 04:16 PM #3 Newbie   Joined: Aug 2009 Posts: 12 Thanks: 0 Re: Algebra... urm.....i dont understand.....
 September 15th, 2009, 05:21 PM #4 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Algebra... List them out: Last digit of 7 = 7 Last digit of 7 * 7 = Last digit of (7 * 7) * 7 = Last digit of (7^3) * 7 = Last digit of (7^4) * 7 = Last digit of (7^5) * 7 = etc. You'll see a pattern to it. If the pattern is, for example (and this is NOT the pattern) 1, 2, 3, 1, 2, 3, 1, 2, 3, ... You can use division to figure out where the 2005th power of 7 would appear on the list. You only need the last digit because the remainder when dividing by 5 only depends on the last digit of a number.
 September 16th, 2009, 07:31 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 1. Or use the fact that 7^2005 = 7((5(480) + 1)^501). 2. Is this worded correctly? N can be 11, which is very obviously its lowest possible positive value, but if negative values are allowed N can be as low as you like. In any case, it's unclear what "respectively" means in this context (for N = 11 to work, a particular interpretation is needed).

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