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21 Nov 2005, 12:18
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nero44
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.
21 Nov 2005, 12:55
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sarojpatra
Yeah, there is very time-saving 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
05 Aug 2009, 11:14
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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.
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.
