official quant thread for cat08

[dashing]

Originally Posted by the_hate

LCM(2,3,4...16) = 2^4*3^2*5*7*11*13
=16*9*5*7*11*13

so no must be of form 16*9*5*7*11*13*k + 1 = 17*m

where k,m are integers

m = (720720k+1)/17
= 42395 + (5k+1)/17
smallest value for k such that (5k+1)/17 is integer is 10
m=42395+3 = 42398

m not confident bout calculation portion

Originally Posted by naga25french (official quant thread for cat08)
New Problem:
Find the least multiple of 17 which when divided by 2,3,4..16 leaves a remainder of 1 in each case

least multiple will be (LCM of 2,3,4.....17 ) +1

[implex]

Originally Posted by dashing

least multiple will be (LCM of 2,3,4.....17 ) +1

when we take lcm of 17 as well, and add 1 to ti
the number won't be divisible by 17 !!

[implex]

Originally Posted by selebratinglife

Posted my solution..You seem to have missed it ..Anyway, reposting
angle ABD=ACD=90 degrees since angle in a semi circle ..And BD bisetc the angle ADC can be proved by alternate segment theorem and ABC being isosceles triangle..From triangles ADB and ADC it can be deduced that CD=3.5

Please explain the part in bold!! how can we deduce that, didn't u deduce cd=1 when the options were not there!!

[deep@k]

Originally Posted by selebratinglife

angle ABD=ACD=90 degrees since angle in a semi circle ..And BD bisetc the angle ADC can be proved by alternate segment theorem and ABC being isosceles triangle..From triangles ADB and ADC it can be deduced that CD=3.5

It is not clear.....
further explanation is required...........

[raghav507]

Originally Posted by implex

yeah it's 3.5 can u post ur solution!!

hi my solution,
have used the formula AC.BD = AB.CD + AD.BC
here CD can be taken as a variable x and the other sides can be found out in terms of x(using the Pythagoras theorem).
it will result in a quadratic equation with roots 7/2 and -4.

hope its clear...in any case i can provide the detailed soln also...

[implex]

Originally Posted by raghav507

hi my solution,
have used the formula AC.BD = AB.CD + AD.BC
here CD can be taken as a variable x and the other sides can be found out in terms of x(using the Pythagoras theorem).
it will result in a quadratic equation with roots 7/2 and -4.

hope its clear...in any case i can provide the detailed soln also...

yupp right, ptolemy's theorem was to be used for the shortest solution. We can also use cosine law!

[naga25french]

Originally Posted by implex

answer is wrong . the concept is simple
it should be k times lcm of(2,3,4..16) +1
N=k.lcm(2,3,4..16) +1
we will find the number k as 10

so the number is N=10.16.9.5.7.11.13+1

well... i made a small mistake in calculating the lcm..

anyways , answer is 7207201

[deep@k]

Originally Posted by raghav507

hi my solution,
have used the formula AC.BD = AB.CD + AD.BC
here CD can be taken as a variable x and the other sides can be found out in terms of x(using the Pythagoras theorem).
it will result in a quadratic equation with roots 7/2 and -4.

hope its clear...in any case i can provide the detailed soln also...

it seems to looking ok....
but if possible please provide detailed soln....

thnx in advance....

[implex]

Originally Posted by deep@k

it seems to looking ok....
but if possible please provide detailed soln....

thnx in advance....

AD is diameter so <ACD=<ABD =90
now in right triangle ABD
BD=Sqrt( 4^2-1^2)=sqrt(15)
in triangle ACD, similarly AC=sqrt(16-x^2) where cd=x

now ptolemy's theorem says , in a convex cyclic quadrilateral ABCD
AB.CD+ AD.BC=AC.BD
we put the values
4+x=sqrt(15)*sqrt(16-x^2)
squaring
x^2+16+8x=240-15x^2
16x^2+8x-224=0
2x^2+x-28=0
solve you will get only one positive root as 7/2

[deep@k]

Originally Posted by implex

AD is diameter so <ACD=<ABD =90
now in right triangle ABD
BD=Sqrt( 4^2-1^2)=sqrt(15)
in triangle ACD, similarly AC=sqrt(16-x^2) where cd=x

now ptolemy's theorem says , in a convex cyclic quadrilateral ABCD
AB.CD+ AD.BC=AC.BD
we put the values
4+x=sqrt(15)*sqrt(16-x^2)
squaring
x^2+16+8x=240-15x^2
16x^2+8x-224=0
2x^2+x-28=0
solve you will get only one positive root as 7/2

this is fine.... excellent piece of work...
but i didnt get that why similar triangle concept is not working here?..

please clarify my doubt......