Find the last two digit in 122*123*125*127*129?
Ans 1)20 2)50 3)30 4)40 5)60
for finding the last 2 digits find the product of expression when divided by 100
i.e.
22*23*25*27*29 / 100
506*675*29 / 100
6*75*29 /100
450*29 / 100
1450/100
thus 50
for finding the last 2 digits find the product of expression when divided by 100
i.e.
22*23*25*27*29 / 100
506*675*29 / 100
6*75*29 /100
450*29 / 100
1450/100
thus 50
153=17*9....128^1000=2^7000.....
Use Chinese remainder theorem........
Rem=Rem=2*-1=-2=7....
Rem=Rem=1.....
9x+18y=1.....x=2 and y=-1 satisfies.....
So Rem=9*2*1+(-1)*17*7=18-119=-101=52.....
Can anybody plz explain Chinese Remainder Theorem with example and where to use it efficiently?
Thanks
its not necesary that B is a single digit integer..
B may b any no. lyk 6, 12, 18, 24, 30..
so it is a positive integer as well as multiple of 6..
ans is d) both a and c..
153=17*9....128^1000=2^7000.....
Use Chinese remainder theorem........
Rem=Rem=2*-1=-2=7....
Rem=Rem=1.....
9x+18y=1.....x=2 and y=-1 satisfies.....
So Rem=9*2*1+(-1)*17*7=18-119=-101=52.....
could you tell me how did you write the equation(highlighted in red))
P.S. beginning Quant. So dont mind with silly doubts...
could you tell me how did you write the equation(highlighted in red))
P.S. beginning Quant. So dont mind with silly doubts...
acc to c.r.t.
if a no.n=a*b (a & b are prime to each other)
and m is a no.such that
remainder=r1 & rem=r2
thn rem=ar2x+br1y=1 where ax+by=1
here 128^1000=m & 153=n=9*17(a=9,b=17)
Q) n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has?
ans a)35 b)32 c)28 d)none of these
Q) n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has?
ans a)35 b)32 c)28 d)none of these
is the ans 35..??
grm_bh Saysis the ans 35..??
yes the answer is correct...what is the solution?
acc to c.r.t.
if a no.n=a*b (a & b are prime to each other)
and m is a no.such that
remainder=r1 & rem=r2
thn rem=ar2x+br1y=1 where ax+by=1
here 128^1000=m & 153=n=9*17(a=9,b=17)
Thanku
But how did br1y become 18y??? b=17 r1=7(-2) so it has to be 119 or -34 i guess.......
rsgage Saysyes the answer is correct...what is the solution?
2n=28=7*4
3n=30=6*5
when n is multiplied by 2 no. of factors may hav bcum 7 frm 6
wen multiplied by 3 no.of factors bcum 5 frm 4...
so original factors=6*4
on multiplication(6n) no of factors=(6+1)*(4+1)=35
Thanku
But how did br1y become 18y??? b=17 r1=7(-2) so it has to be 119 or -34 i guess.......
dats d last step to find the remainder..
b4 dat u hav to takke d eqn ax+by=1
so here..9x+17y should b equal to 1
for this d values which satisfy d eqn r(2,-1)
now subs. these for d next step
(it wont b 18y but 17y)
thanku grm.............
Can anyone pls tell what is the total marks of Cat 2010.pls help
somepne pls post usefull thread??
Can anyone pls tell what is the total marks of Cat 2010.pls help
somepne pls post usefull thread??
Total marks since 2009 has been 180. with 60 in each section. 20 questions in each section carrying 3 marks, negative 1 mark.
for more details please visit
Indian Institutes of Management
:
The sole purpose of thisThread is to bring together the CAT aspirants who are solving the Quant book by Arun Sharma. This will be a common place to discuss any doubts while solving the questions. As there are many questions in the book where either the Hint/Solution is missing or the answer itself is wrong. It will be very nice of you all to help people clear their doubts and also they could learn various methods of problem solving.
:grab:
Thread is to bring together the CAT aspirants who are solving the Quant book by Arun Sharma. This will be a common place to discuss any doubts while solving the questions. As there are many questions in the book where either the Hint/Solution is missing or the answer itself is wrong. It will be very nice of you all to help people clear their doubts and also they could learn various methods of problem solving.hi abhijeet,
this is karun can u help me in solving the problem of numbers lod 2 q no. 97 the mock test conducted by ams learining centre.
plssss
Q) Find all two digit numbers such that the sum of the digits constituting the number is not less than 7; the sum of the square of the digits is not greater than 30; the number consisting of the same digits written in the reverse order is not larger than half of the given number?
ans a) 52 b)51 c)49 d) 53 e)50
Hi..
This is a good way to study..
Q) Suppose the product of n consecutive integer is x.
(x+1)(x+2)(x+3).....(x+(n-1))=1000, then which of the following cannot be true about the number of terms n.
Ans 1. The no. of term can be 16
2. The no. of term can be 5
3. The no. of term can be 25
4. The no. of term can be 20
Hey Puys,
I just started my preparation for CAT 11. I am weak in P and C, probability and Geometry.. would you recommend Arun Sharma for quant for me? As of now I am preparing from CAT 2010s TIME Material.
Thanks
Rv 😁
Q) Suppose the product of n consecutive integer is x.
(x+1)(x+2)(x+3).....(x+(n-1))=1000, then which of the following cannot be true about the number of terms n.
Ans 1. The no. of term can be 16
2. The no. of term can be 5
3. The no. of term can be 25
4. The no. of term can be 20
Its 2......
1000 has three 0's......so the product has to have three multiples of 5.....
Now product of n consecutive integers is x=n!*k......x will have a zero at the end for all n greater than or equal to 5......but we need 0 or 5 at the end of at least three terms among (x+1),(x+2),....(x+(n-1))......the first three terms will be x+5, x+10 and x+15.......so n-1 has to be at least 15.....or n has to be at least 16......
