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• hi plp plz help me with these questions... Q.whats is the remainder on dividing 1) 4^44 by 44 .....answer 36 2) (3^102 + 4^48) by 7....answer 2 Q.if n^n is
Message 1 of 9 , Aug 25, 2005
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hi plp plz help me with these questions...

Q.whats is the remainder on dividing

1) 4^44 by 44  .....answer 36
2) (3^102 + 4^48) by 7....answer 2

Q.if n^n is odd....which of the following is true:

1)n^(n-1)*n^(n+1)is odd integer
2)n^{(n+1)\2)}*n^n^2 is odd integer
here second part is n raised to n raised to 2

3)n^2n+n^(n\2)is even integer

answer.....both 1 and 2 are true

Q.what are the last two digits of 707^444

plzgive the complete solution with explainatio

thanx

Tina

• Hi Tina I think the answer to (4^44)/44 is 0 Explanation: (4^44)/44=((4^4)^11)/44=(256^11)/44=(256/44)*256^10=4*256^10 hence the remainder is 0 the answer to
Message 2 of 9 , Sep 6, 2005
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Hi Tina

I think the answer to (4^44)/44 is 0
Explanation:
(4^44)/44=((4^4)^11)/44=(256^11)/44=(256/44)*256^10=4*256^10
hence the remainder is 0

the answer to (3^102 + 4^48) by 7 is 2
Explanation:
(3^102)/7=(9^51)/7=(2^51)/7=(8^17)/7=(1^17)/7
hence remainder is 1
(4^48)/7=((4^3)^16)/7=(64^16)/7=(1^16)/7
hence the remainder is 1
so toghether we the remainder as 2

Q3) given that n^n is odd which means that n is odd
Which implies that n^(n-1)*n^(n+1) and
n^{(n+1)\2)}*n^n^2 are odd

Q4) 707^444=(7*101)^444=7^444*101^444
101^444 the last two digits will be 01
7^444=(7^4)^101=2401^111.. the last two digits of
this will be 01
so in the product also the last two digits is 01

--- tina <manisha_rai1484@...> wrote:

> hi plp plz help me with these questions...
>
> Q.whats is the remainder on dividing
>
> 1) 4^44 by 44 .....answer 36
> 2) (3^102 + 4^48) by 7....answer 2
>
>
> Q.if n^n is odd....which of the following is true:
>
> 1)n^(n-1)*n^(n+1)is odd integer
> 2)n^{(n+1)\2)}*n^n^2 is odd integer
> here second part is n raised to n raised to 2
>
> 3)n^2n+n^(n\2)is even integer
>
> answer.....both 1 and 2 are true
>
> Q.what are the last two digits of 707^444
>
>
> plzgive the complete solution with explainatio
>
> thanx
>
> Tina

______________________________________________________
http://store.yahoo.com/redcross-donate3/
• Hi Tina, I have solved these questions to the best of my knowledge. Let me know in case of any corrections or alternate solutions. Here are the Answers to your
Message 3 of 9 , Sep 7, 2005
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Hi Tina,

I have solved these questions to the best of my knowledge. Let me know in case of any corrections or alternate solutions.

Soln 1 :

4^44 = (256)^11

256 = 264 - 8

(264 - 8) ^ 11

Now,

(264 - 8) ^ 11 / 44, since 264 is divisible by 44, -8 will be the remainder. As, (-8) ^ 11 will be  taken as -8 because of odd power, we are left with - 8 as remainder.Since, Remainder can't be negative , therefore answer is
- 8 + 44 = 36.

Soln 2 :

3 ^ 102 = (9) ^ 51,  since, 9 = 7 - 2

(9) ^ 51 = (7 - 2) ^ 51 / 7  will give remainder - 2 (odd power);

Similarly remainder can't be negative  therefore it is  - 2 + 7 = 5:

4 ^ 48 = (2) ^ 96 = (8) ^ 32;

8 = 7 - 1; (7 - 1) ^ 32 / 7 wiil give remainder 1;]

therefore overall remainder will be 5 + 1 = 6. However your answer is 2.

Soln 3 :n^n is even or odd depends on the value of n.
Basic rule is (even) ^ n = even { irrespective of n is even or odd}
Similarly,   (odd) ^ n = odd  { irrespective of n is even or odd}

Soln 4:

707 = 101 X 7;

Remember the basic rule , (101) ^ n last two digits are always 01. For instance, (101) ^ 2 = 10201, (101) ^ 3 = 1030301......

Now,

(707) ^ 444 = ( 101 X 7 ) ^ 444 = (101) ^ 444 x (7) ^ 444. Since last digit of 7 ^ 4n is always 1. theerfore last two digits of (707) ^ 444 are 01.

Regards,

Jay

----- Original Message -----
From: tina
Sent: Friday, August 26, 2005 12:56 AM
Subject: [ascent CAT] help

hi plp plz help me with these questions...

Q.whats is the remainder on dividing

1) 4^44 by 44  .....answer 36
2) (3^102 + 4^48) by 7....answer 2

Q.if n^n is odd....which of the following is true:

1)n^(n-1)*n^(n+1)is odd integer
2)n^{(n+1)\2)}*n^n^2 is odd integer
here second part is n raised to n raised to 2

3)n^2n+n^(n\2)is even integer

answer.....both 1 and 2 are true

Q.what are the last two digits of 707^444

plzgive the complete solution with explainatio

thanx

Tina

• Hi, i have read the msg today. i will surely give the answer. plz give explanation of the given problem x^y+y^x=17 & x+y=5. what`s the value of x,y ans x=3,y=2
Message 4 of 9 , Sep 7, 2005
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Hi, i have read the msg today. i will surely give the answer. plz give explanation of the given problem

x^y+y^x=17 & x+y=5.

what`s the value of x,y

ans x=3,y=2  OR x=2, y=3

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• 1) 4^44 by 44 4^4*11 by 44 256^11 by44 36^11 by 44 6^2*11 by 44 6^22 by 44 6^3*7. 6 by 44 216^7 . 6 by 44 40^7 . 6 by 44 40^2*3 . 6. 40 by 44 1600^3 .6.40 by
Message 5 of 9 , Sep 7, 2005
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1) 4^44 by 44

4^4*11 by 44
256^11 by44
36^11 by 44
6^2*11 by 44
6^22 by 44
6^3*7. 6 by 44
216^7 . 6 by 44
40^7 . 6 by 44
40^2*3 . 6. 40 by 44
1600^3 .6.40 by 44
16^3 .6.40 by44
16^2 .6 .40 .16 by 44
256 .6 .40 .16 by44
divide it individually and take the remainder
256 by 44    remainder=36
36 .6 .40 .16 by 44
216 .40 .16 by 44
divide 216 by 44
remainder =40
now
40 .40 .16 by44
1600 .16 by 44
divide 1600 by 44
remainder=16
now
16*16by 44
256by44

2)(3^102 + 4^48) by 7

firstly take 3^102 by 7
3^2*51 by 7
9^51 by 7   on dividing 9 by 7 remainder is 2
now it means
2^51 by 7
2^3*17 by 7
8^17 by 7 divide 8 by 7  remainder is 1
so
1^17=1

now take 2 one

4^48by 7
16^24 by 7
on dividing 16/7
remainder=2
2^24 by 7
8^8 by 7
on dividing
remainder =1
1^8=8

tina <manisha_rai1484@...> wrote:

hi plp plz help me with these questions...

Q.whats is the remainder on dividing

1) 4^44 by 44  .....answer 36
2) (3^102 + 4^48) by 7....answer 2

Q.if n^n is odd....which of the following is true:

1)n^(n-1)*n^(n+1)is odd integer
2)n^{(n+1)\2)}*n^n^2 is odd integer
here second part is n raised to n raised to 2

3)n^2n+n^(n\2)is even integer

answer.....both 1 and 2 are true

Q.what are the last two digits of 707^444

plzgive the complete solution with explainatio

thanx

Tina

Yahoo! India Matrimony: Find your partner online.

Message 6 of 9 , Sep 9, 2005
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hi,

institution called IIPM for MBA in a news paper called
DECAN CHRONICLE.is it worthy to join there?

bala.

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• ... hope u have got the solution by now,if not this might help u Q.whats is the remainder on dividing 1) 4^44 by 44 .....answer 36 2) (3^102 + 4^48) by
Message 7 of 9 , Sep 24, 2005
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--- In ascent4cat@yahoogroups.com, "tina" <manisha_rai1484@r...> wrote:

hope u have got the solution by now,if not this might help u

Q.whats is the remainder on dividing

1) 4^44 by 44 .....answer 36
2) (3^102 + 4^48) by 7....answer 2

sol.

1)4=4(mod 44){its actually equivalent to instead of equal to}
2)16=16(mod 44)
3)64=20(mod 44)
4)256 or 16*16=36(mod 44)=-8(mod 44)
now lets multiply 1 and 4,so
5)4^5=36*4(mod 44) =12(mod 44)
6)4^6=4(mod 44){square up 3}
7)4^7=16(mod 44){6*1}
8)4^8=20(mod 44){7*1}

so now u can observre the repetition thus the answer is 36

u can apply the same for 2nd also and hope if u are clear with this
concept of mod function and power cycle will help u with other
solution also

> hi plp plz help me with these questions...
>
> Q.whats is the remainder on dividing
>
> 1) 4^44 by 44 .....answer 36
> 2) (3^102 + 4^48) by 7....answer 2
>
>
> Q.if n^n is odd....which of the following is true:
>
> 1)n^(n-1)*n^(n+1)is odd integer
> 2)n^{(n+1)\2)}*n^n^2 is odd integer
> here second part is n raised to n raised to 2
>
> 3)n^2n+n^(n\2)is even integer
>
> answer.....both 1 and 2 are true
>
> Q.what are the last two digits of 707^444
>
>
> plzgive the complete solution with explainatio
>
> thanx
>
> Tina
•  Hi, how to solve this : (100+50root2)*3 The answer is 511.5 i want the steps.. Thanks & regards Shyam sawant NSE.iT,Mumbai ltd Hi, how to solve this :
Message 8 of 9 , Dec 3 5:41 AM
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Hi,

how to solve this :
(100+50root2)*3

The answer is 511.5 i want the steps..

Thanks & regards
Shyam sawant
NSE.iT,Mumbai ltd

• This is a simple one Root2=1.41 (You must be knowing this) 1. 50*root2=50*1.41=70.5 2. 70.5+100=170.5 3. 170.5*3=511.5 Got it?? Regards, Sunil Kamat
Message 9 of 9 , Dec 18 10:18 PM
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This is a simple one

Root2=1.41 (You must be knowing this)

1. 50*root2=50*1.41=70.5

2. 70.5+100=170.5

3. 170.5*3=511.5

Got it??

Regards,

Sunil Kamat

Software Engineer | Infosys Technologies Ltd. | Bangalore

Extn: 63245 | Mobile: 91 99451 29390 | Direct: 91 80 515 63245

From: ascent4cat@yahoogroups.com [mailto:ascent4cat@yahoogroups.com] On Behalf Of shyam arvind sawant
Sent: Saturday, December 03, 2005 7:12 PM
To: ascent4cat@yahoogroups.com
Subject: [ascent CAT] help

Hi,

how to solve this :
(100+50root2)*3

The answer is 511.5 i want the steps..

Thanks & regards
Shyam sawant
NSE.iT,Mumbai ltd

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