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I'm going to make a sticky thread where the very very basic

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New post 21 Nov 2005, 12:18
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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New post 21 Nov 2005, 12:55
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 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
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New post 05 Aug 2009, 11:14
3
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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New post 31 Jan 2010, 06:56
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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New post 31 Jan 2010, 11:52
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:  [#permalink]

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New post 09 Jul 2010, 04:39
http://en.wikipedia.org/wiki/Divisibility_rule

Has 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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New post 29 Aug 2019, 02:12
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Re: I'm going to make a sticky thread where the very very basic   [#permalink] 29 Aug 2019, 02:12

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