Need help with these QT questions from Arun Sharma's book

[TheCat]

arre yeah sab kya hai...yeah Cat mein nanhi ata hai..

[indian_girl]

Originally Posted by TheCat

arre yeah sab kya hai...yeah Cat mein nanhi ata hai..

hi , Cat,

Dont know yaar,

was just preparing from Arun sharma's book and this are from its very first chapter and its giving me a complex now

But anyways hope someone can solve them out,

ThanQ

[TheCat]

chalo hum solve karte hein..cat ki khatir..

Q.1 Which of the following can never be in the ending of a perfect square?
a. 6
b. 00
c. 000
d. 1

Ans :- option (c) its pretty clear from option...16,100,121 are perfect square but you can't have a number ending with 000..

Q.2 Which of the following is not a perfect square?

a. 1,00,856
b. 3,25,137
c. 9,45,729
d All of these

Ans: see looking at options b can't be a perfect square as it ends with 7..now check for option A ..its square of 316....so option b is not a perfect square..

Q.3 Find the remainder when 9100 is divided by 8?

a. 1
b. 2
c. 0
d. 4

Ans : do it yourself

-------------------------------------------------------------------

Q. 4 Find the remainder of 21000 when divided by 3?
a. 1
b. 2
c. 4
d. 6

Ans : do it yourself


Q. 5 Find the L.C.M of (x+3) (6x2+5x-4) and (2x2+ 7x +3) (x+3)

a. (2x+1) (x+3) (3x+4)
b. (4x2- 1) ( x+3)2 ( 3x+4)
c. (4x2- 1) ( x+3) (3x+4)
d. ( 2x-1) (x+3) (3x+4)

@indian_girl you have done a mistake i think..it shud be 6x2+5x-4

(x+3)(6x2+5x-4) = (x+3) (2x-1)(3x+4)
(2x2+ 7x +3) (x+3) = (x+3)^2 * (2x+1)

hence LCM = (x+3)^2 * (4x^2 - 1) (3x+4) Ans..

Answer other question will be posted when i get free time..

get back in case of any problem..

[indian_girl]

My mistake ,

I should have checked the questions while posting them,

I have corrected the problems posted above,

Sorry , for having posted problems with errors,

and Thanx for provding solutions for the other questions and making them easier now....

ThanQ

[shweta4188]

hello friends
plz help in these ques of arun sharma of NUMBER SYSTEM
QUES 15......PG 20.........THE GREATEST NUMBER WHICH WILL DIVIDE 4003,4216 AND 4249,LEAVING THE SAME REMAINDER IN EACH CASE.

QUES 23....PG 21.......FIND THE LCM OF 2.5,0.5,0.175

QUES 15........PG 23.FIND THE NUMBER OF DIVISORS OF 1728.
is it the factors which we have to find but by this ans is not matching.

[implex]

QUES 15......PG 20.........THE GREATEST NUMBER WHICH WILL DIVIDE 4003,4216 AND 4249,LEAVING THE SAME REMAINDER IN EACH CASE.

just find the hcf of 4216-4003, 4249-4216 and 4249-4003
it comes to hcf of 33,213,246 clealry the no is 3.. and the remainder is 1

QUES 23....PG 21.......FIND THE LCM OF 2.5,0.5,0.175
lcm of 5/2, 1/2,7/40

lcm is lcm of numerator/hcf of deno= 35/2=17.5

QUES 15........PG 23.FIND THE NUMBER OF DIVISORS OF 1728.
is it the factors which we have to find but by this ans is not matching.
1728=16.108=64.27= 2^6.3^3

no of factors
(1+6)(1+3)=28

[shweta4188]

THANX FOR REPLY

I M HAVING ONE MORE query :
in factors section ........a ques is there :FIND THE SUM AND THE NUMBER OF FACTORS OF 1200 SUCH THAT THE FACTORS ARE DIVISIBLE BY 15.

its ans is found by removing powers of 3 and 5 below 1
and so it is given that the sum of factors is (2^0+2^1+2^2+2^3+2^4)*(5^1+5^2)*(3^1) and consequently no of factors is 10.
but in this sum there r other factors too which r not divisible by 15 like 2^0,2^1.....please xplain this one also

[implex]

Originally Posted by shweta4188

THANX FOR REPLY

I M HAVING ONE MORE query :
in factors section ........a ques is there :FIND THE SUM AND THE NUMBER OF FACTORS OF 1200 SUCH THAT THE FACTORS ARE DIVISIBLE BY 15.

its ans is found by removing powers of 3 and 5 below 1
and so it is given that the sum of factors is (2^0+2^1+2^2+2^3+2^4)*(5^1+5^2)*(3^1) and consequently no of factors is 10.
but in this sum there r other factors too which r not divisible by 15 like 2^0,2^1.....please xplain this one also

find the sum of total factors..
find teh sum of factos not divisble by 15 and subtract

[shweta4188]

cont...

please tell how to find the factors which r not divisible by 15.

ques : find the remainder when 51^203 is divided by 7.
please give a hint.

ques : find the remainder of {(173*261)/13 + (248*249*250)/15}
can we add their respective remainders or we need to multiply......

[shankey_595]

Originally Posted by shweta4188

cont...

please tell how to find the factors which r not divisible by 15.

ques : find the remainder when 51^203 is divided by 7.
please give a hint.

ques : find the remainder of {(173*261)/13 + (248*249*250)/15}
can we add their respective remainders or we need to multiply......

to find out d numbers which r not divisble by 15 are the numbers which are not multiples of both (3 & 5)

whn 51 ^ 203 is divided by 7
remainder whn 51^1 is divided is 2
remainder whn 51^2 is divided is 4
and similarly 51 ^3 is 1
so we can conclude its cyclicity is 3 ie after every three powers remainder repeats itself
like 2,4,1,2,4,1,2,4,1 and so on
now 203 is exactly divisble by 7 so we take d last remainder of cyclicity dat is 1


for this question whn d numbers r multiplied der remainders are also multiplied
whn d numbers r added der remainders r also added
so we have to both add and multiply respectively