Number System

[MavericK]

1..the product of 2 numbers is 7168...and the HCF is 16..find the numbers...

to solve these kind of problems, just follow these 3 steps

1. find the value of product/(HCF^2)
2. find the possible pairs of factors of value obtained in step 1
3. multiply the HCF with the pair of prime factors obtained in step 2

same method was followed by Gautam, and the answers are correct

[iim2003]

Why can't i take the first divisor as 7 and proceed as u say .
In that case I have 7*2 + 4 = 18 ?????

And somehow I am not convinced with the ans as 22 ( as u explained )or 18 ( as above ).
Taking 22 as the ans , the no is of form 35n+22 . Taking n as 1 , i get the no as 57. Now 57 when divided by 7 does NOT leave a remainder of 4.
IMHO the ans should be 32
What say u ???

Originally Posted by anil

answer is right...but here's a short-cut

here d1 = the first divisor = 5
r1 = the first remainder = 2
r2 = the second remainder = 4

hence the reqd remainder = (d1 * r2) + r1
hence (5*4)+2 = 22
HTH

[MavericK]

Originally Posted by iim2003

Why can't i take the first divisor as 7 and proceed as u say .
In that case I have 7*2 + 4 = 18 ?????

huh? u havent read the Qn properly...
it says "Divided Successively"...so then how can u assume 7 as the 1st divisor???? is this clear???

[MavericK]

Originally Posted by iim2003

Taking 22 as the ans , the no is of form 35n+22 . Taking n as 1 , i get the no as 57. Now 57 when divided by 7 does NOT leave a remainder of 4.
IMHO the ans should be 32
What say u ???

the answer sure is 22...this is how it goes..
let A is the number which when divided by 5 gives 2 as the rem...and B is the number which when divided by 7 gives 4 as the remainder...let C be the dividend now..and for simplicity lets assume C to be 1...
hence B = 7*1 + 4 = 11
and A = 5*11 + 2 = 57
now 57%35 = 22...which is the answer!!
is this ok?

[zeus]

hey anil,

great shortcuts.

just got me wondering......U attending CL classes by any chance....somehow u managed to get the values in em questions exactly as someone at CL described em!!!

MOI

[cool_peebs]

Q.What is the remainder when 2000^1000 is divided by 13.
a)1
b)2
c)11
4)3

Q.What is the highest power of 6 that divides 144!
a)70
b)48
c)24
d)69

[zeus]

Originally Posted by cool_peebs

Q.What is the remainder when 2000^1000 is divided by 13.
a)1
b)2
c)11
4)3

2000 could be split up as 2002 - 2, 2002 being a multiple of 13

so we worry bout (-2)^1000 or (2)^1000

split that as 2^4 * (2^6)^6

so we have 16*(64)^6 in nr.

ie 16*(65-1)^6

so the remainder would be decided by 16*1/ 13 => 3 is the remainder

Originally Posted by cool_peebs

Q.What is the highest power of 6 that divides 144!
a)70
b)48
c)24
d)69

144!/2*3

check for the highest power of 3 in 144! as 3 would occour more often than 2 and so the no of pairs making up 6 would depend on the highest power of 3

so we got..

Quotient
144/3 = 48
48/3 = 16
16/3= 5
5/3= 1

sum up the quotients...we get 70.

hth

MOi

[Manpreet]

Hey how do we do this one?

What is the last digit of 22^33^44^55^66^77??

[Absolutely Agent]

22^33^44^55^66^77
can be evaluated by just considering 2 instead of 22 and neglecting higher powers


any power of 33^4n = 3^4n ends in 1 ...that is ...it is of the form 5n+1 thus 2^(5n+1)
as cyclicity of 2 is 5 .....we will get the last digit as 2^1 = 2


HTH

[Chandoo]

Hey how do we do this one?

What is the last digit of 22^33^44^55^66^77??

last digit of 66^77 = 6
last digit of 55^66^77 = 5
last digit of 44^55^66^77 = 44^(something)25 is same as 44^1 = 4
last digit of 33^44^55^66^77=1
last digit of 22^33^44^55^66^77=2

Am i right?

Chandoo