Number System

[Darkness]

Originally Posted by nagi_nov3

i guess the numbers may be 58 and 1 so sum least possible would be 59

Let the HCF of a and b = k
Then the LCM will be product of some number x and the HCF k. therefore LCM = x*k
Given that the difference between the HCF and the LCM is 57.
Therefore.. x*k - k = 57
k*(x-1) = 3*19
Pretty evident that either k =1,3 or 19.
If k=1, x=58 LCM = 58 and (a,b)=(2,29), value of a+b = 31
If k=19, x=4 .. not possible.
If k=3, x=20 LCM = 60. Hence a*b = 180. Possible (a,b) for min(a+b) will be 12,15
Hence min(a+b) = 27.

[vineet.nitd]

since l>=h

hence l-h=57 ---(1)
lh=|ab|

=>l=|ab|/h =xh .....(2)

=>x*h^2=|ab|

from (1) and (2)
xh-h =57
(x-1)h =19*3

x=20 , h =3 => |ab| = 9*20=180

x=4 , h 19 => |ab| = 361 * 4 =1444

clearly 1st case gives the smaller magnitudes of a & b sop we take that case ....
|ab| = 180



so , we can think of minimum value of a+b = -180+(-1)= -181
but it does not satify the criteria for difference of lcm and hcf =57
because a anb b must have to be +ve for this ..

hence we discard this value ..
and many more other values ......which form negative a or b ...


hence its clear that aand b have to be +ve

... minimum value of a+b = 12+15 =27 [a+b are positive here ...]

[chiya]

find the remainder when 1!+2!+3!+.....+49! is divided by 7.

@idiotR u doing number system ,TIME se???

[sunilnatraj]

Originally Posted by chiya

find the remainder when 1!+2!+3!+.....+49! is divided by 7.

@idiotR u doing number system ,TIME se???

Really simple.
All numbers after 6! will have 7 in them so each will give remainder 0. You add the rest and take the remainder which shall give u 5 as the answer

[chiya]

I knew I was missing out on something really obvious!!!!
grrrrrr....
thanks

P.S gosh, i can be sooo dumb!

[anupam will return]

Originally Posted by chiya

I knew I was missing out on something really obvious!!!!
grrrrrr....
thanks

P.S gosh, i can be sooo dumb!

Now do this :

(1! + 2! + 3! + 4! + .....49! )^(49!) when divided by 7 what's the remainder ?

[sunilnatraj]

Originally Posted by anupam will return

Now do this :

(1! + 2! + 3! + 4! + .....49! )^(49!) when divided by 7 what's the remainder ?

Dont u think this could be a tad cumbersome? I mean you would have to take numbers upto 6! repeatedly in sets ranging from 1 to 6. And then evaluate the products and then add. Or am I missing something???

Sunil

[siddhesh_j]

Originally Posted by anupam will return

Now do this :

(1! + 2! + 3! + 4! + .....49! )^(49!) when divided by 7 what's the remainder ?

We know that 1!+2! ... 49! gives remainder of 5.

the remainder series of 5 for 7: is 5,4,2,4,2,4,2 ... i.e 25=5^2 = 4(mod 7) and so on
. now 49! is an even no.So the remainder is 4.

Is my ans right , anupam?if no , plz gimme a pm , n i ll try again n,

if yes heres a superb concept clearing qn from me

Solve for x
x==3 mod 7 ,
x== 2 mod 3,
x== 1 mod 2
x== 4 mod 5( note == means congruent to.)

[anupam will return]

Originally Posted by siddhesh_j

We know that 1!+2! ... 49! gives remainder of 5.

the remainder series of 5 for 7: is 5,4,2,4,2,4,2 ... i.e 25=5^2 = 4(mod 7) and so on
. now 49! is an even no.So the remainder is 4.

Is my ans right , anupam?if no , plz gimme a pm , n i ll try again n,

if yes heres a superb concept clearing qn from me

Siddesh you are partially correct. We know that (1!+2!+....49!) gives a remainder of 5, hence the problem now is :
what's the remainder when 5^(49!) is divided by 7 ?
5^(49!) = (125)^(even number) , 125 divided by 7 yields a remainder of -1.
Hence we have (-1)^(even number) = 1. (answer)

[vineet.nitd]

Originally Posted by anupam will return

Now do this :

(1! + 2! + 3! + 4! + .....49! )^(49!) when divided by 7 what's the remainder ?

its as good as sayin
(1! + 2! + 3! + 4! + .....49! )^(49!) mod(7) = (5)^49! mod(7)

but , (5)^49! mod(7) = (125)^(49!/3)mod(7) = (-1)^even mod(7) =1

hence 1 is the answer ....