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nero44 wrote: Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..
e.g. question:
X is an integer, what is the remainder when it is divided by 10? 1) When divided by 5, the remainder is 2 2). When divided by 2, the remainder is 1
Sure, nero44 I'll put in my share and post some remainder rules
There are several rules concerning remainder theory that you might consider to be useful in dealing with remainder problems. First, there are some rules about dividing odd and even numbers. I' post them.
1. When an odd number is divided by an even number, the remainder will be odd regardless. Irrespective whether the quotient is even or odd, the product of the divisor and divident is even (since when you multiply any number by an even number, the product is even). Therefore, the remainder which is the difference between the product(even) and the dividend (odd) will be odd. Example, when a(odd)/b(even), the remainder will always be odd.
2. When odd number is divided by odd number, the remainder might exist. If the quotient is even, the remainder will inevitably exist. if the quotient is odd, the remainder will be even. And vice versa, if the quotient is even, the remainder will be odd. e. g. 9/5 gives the quotient of 1 (odd) and the remainder of 1 (odd) or 29/7 will give 4 as an even quotient and the remainder of 1 (odd). this is true for any numbers.
3. Even by odd. If the quotient is odd, the remainder will necessary exist, if the quotient is odd, the remainder will be even and vice versa (same as in 2.)
4. Even by even. Remainder will exist irrespective of the quotient.
This maight not be helpful with the problem above, but it can be brilliant help with remainder problems including even/odd. I'll post some more remainder rules later.
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sarojpatra wrote: ywilfred wrote: I'll add on some more over a period of time, but here are for starters:
Divisibility rules:  Know them well
Integer is divisible by: 2  Even integer 3  Sum of digits are divisible by 3 4  Integer is divisible by 2 twice or Last 2 digits are divisible by 4 5  Last digit is 0 or 5 6  Integer is divisbile by 2 AND 3 8  Integer is divisible by 2 three times 9  Sum of digits is divisible by 9 10  Last digit is 0
Some other things to note:  If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.
(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))
 Remember to include '1' if you're asked to count the number of factors a number has While counting the number of factors, we also need to count the number it self as a factor.
Yeah, there is very timesaving method to use when you try to solve a problem with finding the number of factors a specific number can have.
1. You have to write the number as the product of primes as a^p*b^q*c^r, where a, b, and c are prime factors and p,q, and r are their powers.
then, the number of factors the product contains will be expressed by the formula (p+1)(q+1)(r+1). e.g. Find the number of all factors of 1435.
1. 1435 can be expressed as 5^1*17^1*19^1
2. total number of factors of 1435 including 1 and 1435 itself is (1+1)*(1+1)*(1+1)=2*2*2=8 factors
However, the number of ways the number can be expressed as the product of of two other numbers can be found using 1/2*(p+1)(q+1)(r+1)
e.g. In how many ways 1620 can be expressed as the product of two numbers.
1. Express the number as the product of primes. 2^2*3^4*5^1
2. 1/2*(2+1)*(4+1)*(1+1)=15 ways
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banerjeea_98 wrote: Dan wrote: If x > y, then x  y divides x^n  y^n. interesting....can someone provide a proof for this...thx
The root of any polynomial of the form ax^n + bx^(n1) + ..C = 0 is basically that value of x that makes the left hand side 0.
To take an example a simple polynomial
x^2 + 2x + 1 = 0
x + 1 is a root of the polynomial.
Why? Because when you plug x = 1 in the eqn you get 0.
Thus to extend the analogy x^n  y^n will always be zero when x = y. In other words xy will be a root of the polynomial.
QED!



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square roots
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Updated on: 27 Apr 2008, 12:17
According to mathworld, a square root of x is a number r such that r^2=x. In other words, any positive real number has two square roots, one positive, sqrt(x), and one negative, sqrt(x). Although in common usage, square root is generally taken to mean the principle square root, i.e. sqrt(x).
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Originally posted by HongHu on 22 Dec 2006, 14:22.
Last edited by HongHu on 27 Apr 2008, 12:17, edited 2 times in total.



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Re: square roots
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23 Dec 2006, 02:41
HongHu wrote: According to mathworld, a square root of x is a number r such that r^2x. In other words, any positive real number has two square roots, one positive and one negative. Although in common usage, square root is generally taken to mean the principle square root, i.e. sqrt(x).
Sorry.... I must disagree. The function is not reciprocal on the negative values.
Sqrt() always returns a positive number by definition. What is matter is that x must be positive in the sqrt() in order to make the equation existing (at least in the real number set of function).
* Sqrt(4) = 2 cannot = 2.
* Sqrt(9) = 3 cannot = 3.
As well, if we define x = 4, we have sqrt(x) = sqrt(4) = 2.
I join a print of the 2 functions. As we observe, for the power of 2, 2 values of x are possible for 1 value of y. But, for the square root, 1 unique value of x match 1 value of y and moreover, x must be positive.
Attachments
Fig1_Square root function.GIF [ 2.42 KiB  Viewed 3726 times ]
Fig2_power 2 (reciprocal).GIF [ 2.4 KiB  Viewed 3668 times ]



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I agree that sqrt(x) is always positive. However it is also true that x=r^2 have two square roots, one is sqrt(x) and the other is sqrt(x). I've edited the previous post to make it more clear.
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Re: Basic Mathematical Principles:
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05 Aug 2009, 11:14
Here's one :
Divisibility test for 7
For any number: double the units digit and subtract it from the remaining digits. If the resultant number is divisible by 7 then the number is divisible by 7.
To illustrate take : 343 Double the units digit 3  > 6 and subtract it from 34  > 34  6 = 28, which is divisible by 7 therefore 343 is divisible by 7.



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ywilfred wrote: adding more...
Adding on to the odd/even rule honghu wrote...
Adding/subtracting two odds or two evens > even Add/ Subtract an odd and an even > od
Multiplication with at no even number > odd ** Even number in a multiplication will always ensure an even product
Interesting properties:  Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)
E.g. Adding 3,4,5,6 will give a sum with[strike]4[/strike]as a factor Adding 2,3,4 will give a sum with 3 as a factor Basic principles dervied from adding, n, n+1, n+2 etc..
 Adding a consecutive set of odd integers will result in sum that is a multiple of the number of integers
E.g. Adding 2,4,6,8,10 > sum will be multiple of 5 (sum=30) Adding 1,3,5 > sum will be multiple of 3 (sum=9)  An even integer in a multiplication > product divisible by 2  2 even integers in a multiplication > product divisible by 4 etc s=3+4+5+6=18 18 is not divided by 4 where am i wrong?
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sanjayism wrote: ywilfred wrote: adding more...
Adding on to the odd/even rule honghu wrote...
Adding/subtracting two odds or two evens > even Add/ Subtract an odd and an even > od
Multiplication with at no even number > odd ** Even number in a multiplication will always ensure an even product
Interesting properties:  Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)
E.g. Adding 3,4,5,6 will give a sum with[strike]4[/strike]as a factor Adding 2,3,4 will give a sum with 3 as a factor Basic principles dervied from adding, n, n+1, n+2 etc..
 Adding a consecutive set of odd integers will result in sum that is a multiple of the number of integers
E.g. Adding 2,4,6,8,10 > sum will be multiple of 5 (sum=30) Adding 1,3,5 > sum will be multiple of 3 (sum=9)  An even integer in a multiplication > product divisible by 2  2 even integers in a multiplication > product divisible by 4 etc s=3+4+5+6=18 18 is not divided by 4 where am i wrong? The sum of ODD number of consecutive integers is a multiple of number of integers. 2+3+4=9, 3 terms (odd), sum=9 is divisible by 3. The sum of EVEN number of consecutive integers is not a multiple of number of integers. 4+5=9, 2 terms (even), sum=9 is not divisible by 2. Hope it's clear.
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Re: Basic Mathematical Principles:
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09 Jul 2010, 04:39
http://en.wikipedia.org/wiki/Divisibility_ruleHas all the divisibility rules up to 989 lol You wouldn't need to know more than up to 20 in my opinion but I'm going to learn up to 20  the less time it takes to work out this stuff on the exam the better. E.g. when you get a large number like 221  ... which is divisible by 17, the rule for divisibility by 17 is: "Subtract 5 times the last digit from the rest. 221: 22  (1 × 5) = 17." What is everyone else's opinion on this  know divisibility rules for numbers up to and including 20 or only up until 11 or 12?



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Re: Basic Mathematical Principles:
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01 Oct 2010, 11:23
The sum of N consecutive integers: S=(n+n+N1)*N/2=n*N+N*(N1)/2 S/N=n+(N1)/2 If N is odd, then S would be a multiple of N. If N is even, then S would not be a multiple of N. i don't understand this.....
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Re: Basic Mathematical Principles:
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12 Nov 2010, 20:08
shrive555 wrote: The sum of N consecutive integers: S=(n+n+N1)*N/2=n*N+N*(N1)/2 S/N=n+(N1)/2 If N is odd, then S would be a multiple of N. If N is even, then S would not be a multiple of N. i don't understand this..... Suppose you are adding up N consecutive integers from n+1 to m. In other words, mn=N, or, m=N+n. And let the sum be S. Now, let's say S1=1+2+...+n S2=1+2+..+n+(n+1)+...+m Then S=S2S1 You know that S1=n(n+1)/2 S2=m(m+1)/2 Therefore S=(m(m+1)n(n+1))/2 =(m^2n^2+mn)/2 =(mn)(m+n+1)/2 =N(N+2n+1)/2 =(N^2+2nN+N)/2 =N(N+1)/2+nN therefore, S/N=(N+1)/2+n We want to see if S can be divisible by N. If N is odd, then N+1 is divisible by 2 and S/N is an integer. In other words, S is divisible by N, or S is a multiple of N. If N is even, then S/N is not an integer. S is not a multiple of N.
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Re: I'm going to make a sticky thread where the very very basic
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Re: I'm going to make a sticky thread where the very very basic &nbs
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