I Am Starting This New thread to help other and solidfy my own Fundamental In maths We Will start with some basic Question In maths which Most of us found them difficult to Handle We will Concentrate mainly on Number System , Fractions…
I Am Starting This New thread to help other and solidfy my own Fundamental In maths
We Will start with some basic Question In maths which Most of us found them difficult to Handle
We will Concentrate mainly on Number System , Fractions and percentage ,Ratio and proportion ,Time Speed, Distance and work, Algebra, Geometry. some of propability and permutation and combination
I Hope will get Support from the best Brain in this forum.
Lets take 1st case
When (2)^256 is divided by 17 What is the remainder ?
We Will make is easy by the way of 256= 8 * 32
Hence 2^256 = ]^32 mean we can write 2^8 32 times
Now 2^8 = 256 Now 256/17 will give us remainder 1 Hence the remainder of 2^256 is also 1
It looks Complicated initially But if we break it in smaller terms we get the solution
Ok Now Ask you One Question
What is the Remainder you get when you divide 5^125 by 124 ?
Please be easy in approach for every one to understand.
Ok Now Ask you One Question
What is the Remainder you get when you divide 5^125 by 124 ?
Please be easy in approach for every one to understand.
The ans is 25. U can have multiple approaches for this ques but the quickest one is that break 5^125 as 5^123x25. Now 5^123= 125^41 which wil give remaindet as 1 hence the ans is 25.
One thing i would like to suggest is that lets share and discuss all the shortcuts available for the main topics of quant bcoz i feel now its the best time to just revise the tips and techniques and learn some good shortcuts you never know what might strike there.
Ok Now Ask you One Question
What is the Remainder you get when you divide 5^125 by 124 ?
Please be easy in approach for every one to understand.
the anser is 25..
Sol:- 5^125= 5^2 * 5^123= 5^2 * (5^3)^41
hence wen (5^125)/124 = /124 = 25 * (1)^41 =25.
Now lets us try :
16^5 + 2^15 Divisible By:
I am not giving the choice at it will ease the problem , Please give the lucid answers.
One thing i would like to suggest is that lets share and discuss all the shortcuts available for the main topics of quant bcoz i feel now its the best time to just revise the tips and techniques and learn some good shortcuts you never know what might strike there.
That is infact A good idea, We must Start with Final Rivision of the Fundamental
Fundamentral of Number System :
I Understand many people have problem Finding The HCF or LCM
Let me share my learing in Lucid Way.
HCF : It is Nothing but the lowest Powered factor which is Common to all numbers
E.g :
150 : 5*5*3*2
210 : 5*2*7*3
375 : 5*5*5*3
here we can easily make out that 5 and 3 is common in all.
hence HCF : 5 * 3 =15
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LCM : It is Nothing but the Highest Powered factor of prime No in all, But may Not be Common.
If we have to find the LCM of the Numbers
150 : 5*5*3*2 ==== 5^2 * 3 * 2
210 : 5*2*7*3 ==== 5*2*7*3
375 : 5*5*5*3 ==== 5^3 * 3
Her we can see that max powered no is 5 is 3 --5^3
and 3 is 1
, 2 is 1
, 7 is 1
so the LCM will be : 5^3 *2*3*7 = 5250
remember HCF(N1,N2).LCM(N1,N2) = N1.N2
1 Like
Some Theorems of Divisibility
1. The Highest power of a prime No p is which divide N! exactly is
N/p + N/(p^2) +N/(p^3)+ ..........
Depending Upon the value of N, if N> p^2 then we will have
N/p + N/(p^2)
Let us take an example
What is the Highest power of 3 which will divide 80! ?
Here 3^= 9, 3^3 = 27 , 3^4 = 81
Here we have
N/p + N/(p^2) +N/(p^3) = 80/3+ 80/9 +80/27 = 26 + 8 + 2 here we take only the Integer
1 Like
Now lets us try :
16^5 + 2^15 Divisible By:
I am not giving the choice at it will ease the problem , Please give the lucid answers.
The Answer to This Question is As below
we first make them to the power of 2
2^4=16 , hence 2^20 + 2^15 = 2^15(2^5 +1) = 2^15(33) hence it is divisible by 33
1 Like
That is infact A good idea, We must Start with Final Rivision of the Fundamental
Fundamentral of Number System :
I Understand many people have problem Finding The HCF or LCM
Let me share my learing in Lucid Way.
HCF : It is Nothing but the lowest Powered factor which is Common to all numbers
E.g :
150 : 5*5*3*2
210 :5*2*7*3
375 : 5*5*5*3
here we can easily make out that 5 and 3 is common in all.
hence HCF : 5 * 3 =15
---------------------------------------------------------------
LCM : It is Nothing but the Highest Powered factor of prime No in all, But may Not be Common.
If we have to find the LCM of the Numbers
150 : 5*5*3*2 ==== 5^2 * 3 * 2
210 : 5*2*7*3 ==== 5*2*7*3
375 : 5*5*5*3 ==== 5^3 * 3
Her we can see that max powered no is 5 is 3 --5^3
and 3 is 1
, 2 is 1
, 7 is 1
so the LCM will be : 5^3 *2*3*7 = 5250
remember HCF(N1,N2).LCM(N1,N2) = N1.N2
gr8 initiative buddy...!!! I am sure this will help us sharpen our fundas just bfor D-DAY..!!!
Happy Dasherra
Another great thread which is gonna be reall helpful for those who were snooring till now.
This looks like an LR puzzle but surely will land-up on QA section.
A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.
Is there a definite approach to it rather than trying all the combinations?
1 Like
A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.
Did not find any Good Arroach But have some good calculation
it is like that
a+b+c+d=17
d+e+f+g=17
g+h+i=17
Start with G+h+i taking any number say 9+7+1 we will decide G later
G can take either 9,7,1 we make D+e+f can take 8,10,16, we cannot take 9,7 as our sum will require shorter no only. so we take 1 as G
so D+E+F = 16 , 8+5+3 But we have not decided what D can take so we first add A+B+C we are left out with number 6 and 2 as A=4 known so 6+4+2 = 12 D can take only 17-12 =5
Hence D,G can take 5,1
another easy approach (Striked immediatly after i have solved the problem.) is
we have 1+2+3+4+5+6+7+8+9 = 45 any 2 number are repeated
a+b+c+d=17
d+e+f+g=17
g+h+i=17
which make a total of 51 so the addional no require is (17*3)-45 = 6
combination can be only (1,5) , (2,4) sicne A=4 so we left with only choice (5,1) -- (1,5) Not possible try . No need to solved as i did. but anyway keeping as what approach i used to help other understand.
Did not find any Good Arroach But have some good calculation
it is like that
a+b+c+d=17
d+e+f+g=17
g+h+i=17
Start with G+h+i taking any number say 9+7+1 we will decide G later
G can take either 9,7,1 we make D+e+f can take 8,10,16, we cannot take 9,7 as our sum will require shorter no only. so we take 1 as G
so D+E+F = 16 , 8+5+3 But we have not decided what D can take so we first add A+B+C we are left out with number 6 and 2 as A=4 known so 6+4+2 = 12 D can take only 17-12 =5
Hence D,G can take 5,1
another easy approach (Striked immediatly after i have solved the problem.) is
we have 1+2+3+4+5+6+7+8+9 = 45 any 2 number are repeated
a+b+c+d=17
d+e+f+g=17
g+h+i=17
which make a total of 51 so the addional no require is (17*3)-45 = 6
combination can be only (1,5) , (2,4) sicne A=4 so we left with only choice (5,1) -- (1,5) Not possible try . No need to solved as i did. but anyway keeping as what approach i used to help other understand.
Great approach man really. Keep up the good work. One ques from my side
if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?
The Answer to This Question is As below
we first make them to the power of 2
2^4=16 , hence 2^20 + 2^15 = 2^15(2^5 +1) = 2^15(31) hence it is divisible by 31
is tht 33.
i mean i got it 33 yaar, hey bro u made a mistake in addition:) ....
anyways u guys r doing a gr88 job by introducing this thread,i mean its a kind of
gr88 help to ppl like me who r pathetic in quant.i too shall contribute to make it successful. cheers.......
Another great thread which is gonna be reall helpful for those who were snooring till now.
This looks like an LR puzzle but surely will land-up on QA section.
A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.
Is there a definite approach to it rather than trying all the combinations?
add alll we get 45+ D+G =51
d+G= 51 = 1,5 or 2,4 or 3,3
3,3 and 2,4 not possible
so 1,5
E+F= 11
also if G= 1 H+I= 16
now if G= 5 H,I =9,3
E,F can be 6,5 7,4 8,3 9,2 but none is possible so G=1 D=5
if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?
Can Any one put some easy light for us to see how we can get the solution for such questions
Awesome thread!! Just what we need to brush up our basics!!
if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?
Well my QA is worse but i think I can give this one a try... PLz correct me if i m wroung or some one finds a easy way out....
WELL x + y + z = 10
My approch is based on assumption and susbtitution..
asuming X=0
y+z=10
for this y & z can have 10 st of values from (0,10) to (10,0)
similarly assuming X=1
y+z=9 ===> y,z have 9 pair of values
Going by this approach
total no. of values for (x,y,z) = 10+9+8+7+6+5+4+3+2+1=55
So i think 55 is the answer.
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NEVER SAY DIE...........:smilecol: :snipersm:
Well my QA is worse but i think I can give this one a try... PLz correct me if i m wroung or some one finds a easy way out....
WELL x + y + z = 10
My approch is based on assumption and susbtitution..
asuming X=0
y+z=10
for this y & z can have 10 st of values from (0,10) to (10,0)
similarly assuming X=1
y+z=9 ===> y,z have 9 pair of values
Going by this approach
total no. of values for (x,y,z) = 10+9+8+7+6+5+4+3+2+1=55
So i think 55 is the answer.
if x,y and z are real numbers then there could be infinite solutions.However, if x,y and z are positive integers then i am getting 66 as answer.plz correct me if i am wrong.
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