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Showing posts with label factors. Show all posts
Showing posts with label factors. Show all posts
Monday, June 27, 2016
Monday, February 8, 2016
Puzzle
There are a 1000 locks in closed position. 1000 children numbering 1 to 1000 approach the lock one by one. The rule is the child (say with number 'n') can open or close the lock numbering in multiples of "n" only.
so the child in number one position can open all the locks. Child number 2 will be able to close all the locks in multiples of 2 (like 2, 4, 6 .... upto 1000). The other locks will remain as they are.
After all the 1000 children have had their chance, how many locks will be in open position.
so the child in number one position can open all the locks. Child number 2 will be able to close all the locks in multiples of 2 (like 2, 4, 6 .... upto 1000). The other locks will remain as they are.
After all the 1000 children have had their chance, how many locks will be in open position.
Tuesday, December 30, 2014
Monday, December 29, 2014
Friday, October 3, 2014
factors
For the number 610 x 717 x 5527,
find
1.
The number of prime factors
2.
The number of all the factors.
tutorials -sum of factors
Sum of Factors
Another Example:
If a number N can be expressed as product of prime factors such that
N = pa x qb x rc...............................
Write the power of each of the prime factors beginning with zero and going to thee highest power of the prime number.
Record the same for each prime number.
An example will help
64= 26
Then the sum of factors (20 + 21 + 22
+ 23 + 24 + 25 + 26) = 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127
Another Example:
Find the total number of factors of 222.
222 = 2 x 3 x 37
Or, 222 = 21 x 31 x 371
then the sum of factors are (20
+ 21) (30 + 31) (370 + 371) = (1+2)(1+3)(1+37) = 3 x 4 x 38 = 456
We will check the answer
The factors of 222 are 1, 2, 3, 6, 37, 74, 111, 222.
Sum of factors = 456
tutorial on number of factors
To find the number of factors of a
number
If a number N can be expressed as product of prime factors
such that
N = pa x qb x rc...............................
Then, the total number of factors including N and 1 is given
by
(a + 1)x(b + 1)x(c + 1).................................
Example:
Find the total number of factors of 222.
222 = 2 x 3 x 37
Or, 222 = 21 x 31 x 371
So, the total number of factors = (1+1)x(1+1)x(1+1) = 2x2x2
= 8
We can check the answer as shown below.
The factors of 222 are 1, 2, 3, 6, 37, 74, 111, 222.
Find the total number of factors of 64.
64 = 26
So, the total number of factors = (6+1) = 7
We can check the answer as shown below.
The factors of 64 are 1, 2, 4, 8, 16, 32, 64.
Thursday, September 25, 2014
JMO
The least common multiple of two natural numbers a and b, is 399. What is the minimum possible sum of the digits of the number a(given a > b)?
Wednesday, August 6, 2014
Thursday, August 1, 2013
solution to question paper 1 - IPM STD V
The least number to be multiplied by 17640 so that the
result becomes a perfect square is -------------
Answer:
17640 = 3 x 3 x 2 x 2 x 7 x 7 x 2 x 5
So, the least that should be multiplied = 2 x 5 = 10 to make
it a perfect square.
Thursday, July 11, 2013
factors trick 2
Say you have a number 64.
It can be written as 2 x 2 x 2 x 2 x 2 x 2 x 1.
The no. of factor will simply be the count of all the factors.
ie. 7
It can be written as 2 x 2 x 2 x 2 x 2 x 2 x 1.
The no. of factor will simply be the count of all the factors.
ie. 7
number of factors trick 1
consider the number 210
it can be written as 210 = 2 x 5 x 3 x 7
All the factors are prime, and all the prime factors occur only once.
in such a case, the total number of factors is 2 x 2 x 2 x 2. (as many 2s' as there are prime factors).
So, the answer is 16.
consider 105.
it can be written as 105 = 3 x 5 x 7
All the factors are prime, and all the prime factors occur only once.
in such a case, the total number of factors is 2 x 2 x 2 (as many 2s' as there are prime factors).
So, the answer is 8.
it can be written as 210 = 2 x 5 x 3 x 7
All the factors are prime, and all the prime factors occur only once.
in such a case, the total number of factors is 2 x 2 x 2 x 2. (as many 2s' as there are prime factors).
So, the answer is 16.
consider 105.
it can be written as 105 = 3 x 5 x 7
All the factors are prime, and all the prime factors occur only once.
in such a case, the total number of factors is 2 x 2 x 2 (as many 2s' as there are prime factors).
So, the answer is 8.
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