You have a product of five numbers. For this product is divisible by 5 simply one factor to be. We

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Hello, I'm on a chapter divisibilitequi does not look very complicated but I'm stuck on Dtails So here is an example on how I blocks: Determine the integers n such that 2n +3 divides the four dividers are D 4 (4) = -4, -2, -1, 1, 2, 4-2n = 3 -4 n = -7 / 2 -2 n = 2n +3 = -5 / 2 = 2n -1 3 -2 n = 3 2 n = 1 n = 2 n -1 = 2 n = 3 -1 / 2 4 n = 2 n = 3 1/2 searches Whole Here are -2 and -1 what I do not understand is that 2n +3 divides 4 therefore 2n +3 | 4 therefore 4 = k * (2n +3) so I have to find k. Now k = -4 or -2 or -1 or 1 or 2 or 4, but in the example my teacher gave me, it was not as procde, of what I see after it has 2n +3 = k and I understand why it is not his .... thank you in advance for your help!
Hi, It would be good it is prciser integers. For my part when it comes to whole it is natural integers. stack on safe In terms of your question, it is possible that your teacher note k dividers 4 and it falls on the same values as you.
^ ^ No, find the values come from the teacher, me what I do not understand is that 2n +3 divides 4 therefore 2n +3 | 4 therefore 4 = k * (2n +3) so I have to look k. Now k = -4 or -2 or -1 or 1 or 2 or 4, but in this example, the teacher made this equation: 2n-3 = -4 n = -7 / 2 but why he made his, I understand not its Methodius ... we're supposed to look n That's how I PROCD: for k = -4 -4 = 4 (2n +3) =-8n -12 n = (4 +12) / -8 = -2 k = -2 -2 4 = (2n +3) n = -2.5 for k = -1 4 = -1 * (2n +3) n = -7 / 2 for k = April 1 = 2n +3 n = 1/2 k = 2 4 = 2 (2n +3) n = -1 / 2 k = 4 4 = 4 (2n +3) n = -1 I found out the same values but it is Methodius my teacher that I do not understand ....
On 18-11-2012 0:28 @ phis: but in the example, the teacher made this equation: 2n + 3 = -4 n = -7 / 2 This is what I understand: your prof called k dividers 4. It has to be k = -4 and -2 and -1 and 1 then 2 then 4. This enables it to calculate n by the equality stack on safe k = 2n +3
but how this equality can find n? a | b so b = k * a we apply this formula to find n, but after what I see in the example, a | b, so it is to find n b = k but he fate of this or equality?
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ok! if I have another year if you could correct me would be cool! N is an integer, we set f (n) = (11-n) (n-22) (n-33) (n-44) (n-55) establish that f (n) is a multiple of 5 any integer n is what I do: n Z must ensure that f (n) is a multiple of 5 so f (n) = n so 5k-55 = 5 n = 55 n = 5 + 5 (e + 11) = 5k n 'k' = 11 therefore stack on safe e f (n) = (11-n) (n-22) (n-33) (n-44) * 5k 'Let j = (n -11) (22-n) stack on safe (n-33) (n-44), therefore f (n) = j * 5k 'f (n) = 5 (k + j / 5) = 5k with k = k' * j / 5 but its not seem very fair to me all his ...
The fact that f (n) is a product of 5 suggre factors that one of these factors must be divisible by n 5 whatever. As dmontre that when n (n +1) (n +2) is divisible by 3. What remains of 11-n, 22-n, 33-n, 44-n, 55-n when divided stack on safe by 5?? We put n = 5q + r.
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You have a product of five numbers. For this product is divisible by 5 simply one factor to be. We must therefore look at what are the remnants of these 5 numbers when divided by 5. If n = n then 5q + r-11-10 = 5q + r-1 = 5 (q-2) + r n-1-22-20 = 5q + r 2 = 5-(4-q) + r- 2 ....... n-55 = 50 + 5q-r-5 = 5 (q-10) + r-5 You just have to show that one of the residues is zero gal. This is obvious when you considres stack on safe possible remainders of the division by 5.
6:56 p.m. on 18-11-2012 #
When you shout n = 5q + r suppose you do nothing else on n r is the remainder when divided by 5 thou. r can take (when n belongs to N) values 0, 1, 2, 3, 4. So when r belongs to {0, 1, 2, 3, 4} just look what the rest of the division by 5 n-11 n-22, ..., n-55. This remainder is forcment 0, 1, ..., 4. (I'm working on N, you can extrapolate Z). You see if the rest of the division of n by r 5 is the remainder of the division of n-11 r-1 5 is the remainder of the division of n-22 r-5 is 2 .... The 5 remains therefore traverse the set {0, 1, 2, 3, 4} ===> there is therefore one that is zero. (Altered by the Ali_Gator 18-11-2012 7:02 p.m.)
olala is really galre .... When you shout n = 5q + r suppose you do nothing else on n r is the remainder when divided by 5 thou. r can take (when n belongs to N) values 0, 1, 2, 3, 4. so far I understand! therefore n = n = 5k or 5k 5k +1 or n = 2 or n = 3 or n = 5k 5k +4 then you told me to look for the rest of the division by 5 n-11. .., n-55 n-55 so 5k = -55 (I just replace n by 5k) n - 55 = 5 (k-11) n 11 = 5k-1-11 n-11 = -10 = 5k