plz help - number systems

apgo2000 · Joined: 13 Jul 2006 Posts: 15

1) find gcd of (2^100 - 1), (2^120 - 1)

2) find gcd of (1111 ...... 11 hundred ones,
1111...11 sixty ones)

3) 32^32^32 will leave what remainder when divided by 9

4) which is greater:
200^300 or 300^200 or 400^150

5) the sides of a pentagonal field (not regular) are 1737 m, 2160 m, 1422 m, and 2214 m. find the gratest length of the tape by which the five sides may be measured completely:
a) 7 b) 13m c) 11 d)9

6) find no of zeroes in product 1^1 * 2^2 * 3^3 * ....... *99^99 * 100^100

7) find last two digits of 15* 37* 63 * 51 * 91 * 17

8) a bag contains a total of 105 coins of Rs. 1, 50 p and 25 p denominations. find the total number of coins of Re. 1 if there are a total of 50.5 rupeees in the bag and it is known taht the number of 25 paise coins are 133.33 % more than the number of 1 rupee coins.

plz if anyone of u can try these questions plz do tell me the answer and how u did it and if there is any shortcut for such questions. thanx in advance

regards,
anshu

dass_07 · Joined: 02 Mar 2006 Posts: 9

hi,
the answer for q5 is 9

apgo2000 · Joined: 13 Jul 2006 Posts: 15

quote="dass_07"]hi,
the answer for q5 is 9[/quote]

ya u r right. cud u plz tell me how u did it. i think it is a question of hcf but i just cant get the factors right.

and plz if anyone cud try the other questions too.

regards,
anshu

apgo2000 · Joined: 13 Jul 2006 Posts: 15

someone plz help

i am not getting the answers to the questions i have posted
wake up guys

regards,
anshu

P.S. and plz wen u reply give the logic also. i have the answers. i wanna know how u guys got the answer.

palagg · Joined: 22 Feb 2006 Posts: 12

1.the answer to your 5th ques is 9……ya it is done by finding HCF..
1737- 3^2*193
2160- 2^4*5^1*3^3
1422- 2^1*3^2*79
2214- 2^1*3^3*41
as u can c the HCF comes out to b 3^2= 9 hence the answer 9

2.is the answer to your 7th ques is 05??

I will try to solve other probs soon and send u the answers..

Cya
Take care

praphullab · Joined: 18 Sep 2006 Posts: 2

3) 32^32^32 will leave what remainder when divided by 9

taken x= 2 then 9 is x^3 + 1 and 32^32^32 is (x^5)^(x^5)^(x^5)

to find the reminder we need to substitue the solution of x^3+1=0
so we get the remonder by substituting x=-1 in the equation which is -1 so the remonder will be 9-1=8

4) which is greater:
200^300 or 300^200 or 400^150

400^150 = (2*200)^150=(2^150)*(200^150) < 200^300

200^300 = (100^300)*(2^300)
=(100^300)*(8^100)
=(100^200)*(800^100)

300^200 = (3^200)*(100^200)
=(100^200)*(9^100) <100> 200^300 is greater than the other two.

6) find no of zeroes in product 1^1 * 2^2 * 3^3 * ....... *99^99 * 100^100

you need to find the power of 10 to find the no of '0' so we need to find the power of 2*5

as power of 5 will be less than power of 2, power of 10= power of 5

power of 5 in given expression is 5+10+15+....100+25+50+75+100=1300

There will be 1300 zeros in the given expression.

Cool

a bag contains a total of 105 coins of Rs. 1, 50 p and 25 p denominations. find the total number of coins of Re. 1 if there are a total of 50.5 rupeees in the bag and it is known taht the number of 25 paise coins are 133.33 % more than the number of 1 rupee coins.

let the no of 1 rp coins be 12x then 25ps coins will be 28x so the 50ps coins will be 105-40x

12x+(0.5*(105-40x))+(0.25*28x)=50.5

solving for x we get x as 2 so the no of 1 rs coins is 24.

praphullab · Joined: 18 Sep 2006 Posts: 2

2) find gcd of (1111 ...... 11 hundred ones,
1111...11 sixty ones)

GCD of 100 and 60 is 20 so GCD of given two numbers is 111...11 twenty ones