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My humble contribution. Apologies if some of them are re post.
1. A prime number > 5 is always either of the form 6k +1 or 6k -1
2. Product of n consecutive number is always divisible bu n!
3. Number of factors of a number expressed by N = a^x * b^y * c^z ......
is equal to (x+1)*(y+1)*(z+1).....
where a,b,c are prime factors
[This is important]
A perfect square such as 4,9,16 will always have odd number of factors.
Why??? think about it... all prime factors of a square number will always have
even power. And even + 1 = odd. so,using point 3 from above (odd * odd * odd)
is always odd. I have come across many questions that have tested this concept.
Ex. Number of factors of 3528
3528 = 2^3 * 3^2 * 7^2
so number of factors = (3+1)*(2+1)*(2+1) = 4*3*3 = 36
4. A square of an integer which is not divisible by 3 is of the form 3K + 1.
5. An series of number in Arithmetic progression will always have median = mean
6. Number ending with:
a. one zero --- always div by 5
b. two zero --- divisible by 5^2 = 25
c. three zeroes ---- divisible by 5^3 = 125
and so on...
7. 5^n n>2
a. ends with 125 ---- if n is odd
b. ends with 625 ---- if n is even
8. For a given set of numbers:
Arithmetic mean >= geometric mean
Ex: consider three numbers a,b,c
a+b > 2sqrt(ab)
9. To find out the higest power for a given prime factor in a (number)! , consecutively divide the number by that prime
factor and keep on adding the quotient.
For Ex: Find the number of zeroes in 26!
Number of trailing zeroes can be found out if we can find out the highest power of 10.
Each multiple of 5 will contribute a 5 that will multiply with a multiple of 2 to give a 10.
Since multiples of 5 will always be less than multiples of 2 we need to effectively find out the highest power of 5.
so consecutively divide 26 by 5 and add the quoteint.
26mod5 = 5,1 (quotient, remainder)
5mod5 = 1,0 (quotient, remainder)
Adding the quotient ... 5+1 = 6
So highest power of 5 is 6 and hence there will be 6 zeroes at the end in 26!.
10. For n consecutive integers :
Sum is divisible by n if n is odd.
Sum is not divisible by n if n is even.
Will be posting some more facts soon...
1. A prime number > 5 is always either of the form 6k +1 or 6k -1
2. Product of n consecutive number is always divisible bu n!
3. Number of factors of a number expressed by N = a^x * b^y * c^z ......
is equal to (x+1)*(y+1)*(z+1).....
where a,b,c are prime factors
[This is important]
A perfect square such as 4,9,16 will always have odd number of factors.
Why??? think about it... all prime factors of a square number will always have
even power. And even + 1 = odd. so,using point 3 from above (odd * odd * odd)
is always odd. I have come across many questions that have tested this concept.
Ex. Number of factors of 3528
3528 = 2^3 * 3^2 * 7^2
so number of factors = (3+1)*(2+1)*(2+1) = 4*3*3 = 36
4. A square of an integer which is not divisible by 3 is of the form 3K + 1.
5. An series of number in Arithmetic progression will always have median = mean
6. Number ending with:
a. one zero --- always div by 5
b. two zero --- divisible by 5^2 = 25
c. three zeroes ---- divisible by 5^3 = 125
and so on...
7. 5^n n>2
a. ends with 125 ---- if n is odd
b. ends with 625 ---- if n is even
8. For a given set of numbers:
Arithmetic mean >= geometric mean
Ex: consider three numbers a,b,c
a+b > 2sqrt(ab)
9. To find out the higest power for a given prime factor in a (number)! , consecutively divide the number by that prime
factor and keep on adding the quotient.
For Ex: Find the number of zeroes in 26!
Number of trailing zeroes can be found out if we can find out the highest power of 10.
Each multiple of 5 will contribute a 5 that will multiply with a multiple of 2 to give a 10.
Since multiples of 5 will always be less than multiples of 2 we need to effectively find out the highest power of 5.
so consecutively divide 26 by 5 and add the quoteint.
26mod5 = 5,1 (quotient, remainder)
5mod5 = 1,0 (quotient, remainder)
Adding the quotient ... 5+1 = 6
So highest power of 5 is 6 and hence there will be 6 zeroes at the end in 26!.
10. For n consecutive integers :
Sum is divisible by n if n is odd.
Sum is not divisible by n if n is even.
Will be posting some more facts soon...
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08 Sep 2009, 17:12
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urchin
It's a good list of facts, but this last one is not true. If numbers are in arithmetic progression (i.e. if they are 'equally spaced'), it's certainly true that the median of these numbers equals the mean of these numbers - that part is right. However, there are many sets of numbers (any 'symmetric' set, for example) which have an equal median and mean, but which are not in arithmetic progression. For example, both of the following:
0,1,5,9,10
and
4,5,5,5,5,5,6
have a mean and median of 5, but are not in arithmetic progression.
08 Sep 2009, 22:58
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Yes Ian, i guess u are right. Thanks for correcting me. I think the above rule applied only for a set of three numbers.
Because for a set of three numbers, a symmetric set is as good as a arithmetic progression
Because for a set of three numbers, a symmetric set is as good as a arithmetic progression
