About this discussion

- © 2015-2016
- Careers
- Advertise
- Contact Us

Please continue here with all the quant queries and their discussions. link to previous thread :- http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html :cheerio: :cheerio: :c...

Follow this discussion to get notified of latest updates.

Nice! Share it on Facebook so that your friends can join you here.

Order by:
New Posts

Page 52 of 915

If we add the square of the digit in the tens place of the positive two digit number to the product of the digits of that number we get 52,and if we add the square of the digit in the unit's place to the same product of the digits we get 117.Find the two digit number??

i got till this step

x^2+xy=52

x(x+y)=52--(1)

y^2+xy=117

y(y+x)=117--(2)

after this hw to proceed the given answer is 30..pls explain in details with steps.

thanks in advance

10x+y is the number

x*y+x^2 =52

x*y+y^2 =117

y^2-x^2 = 65

y+x = 13

y-x= 5

aading 2y = 18 y= 9

x= 4

so value is 49

Ani1308 Sayshi..a little help in remainder part...find the rem of (3^5^7^9)/41

5^7^9 = 5k

so it is 4k+1

3*3*3*3*3 /41 remainder 38

Commenting on this post has been disabled.

jain4444 SaysThe sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

P.S.:- jain sahab itna easy kaise de sakte hai

Commenting on this post has been disabled.

Varun Tyagi
@varun.tyagi
14,220

hi sry man...i dont have the answers bt i have one more question on the seme pattern...if u liketo practice dn..

5^25^125^3125 /11..

thanks all...

The remainder is

25^anything has unit digit 5

5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1

Commenting on this post has been disabled.

Ani Rai
@Ani1308
349

5^25^125^3125 /11..

thanks all...

Commenting on this post has been disabled.

Varun Tyagi
@varun.tyagi
14,220

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1

=> 5^(16K+7) % 40 = 5

3^(40K + 5) % 41 = (3*81) % 41 = -3 =38

The remainder is38

Ani1308 Saysbut dude u cant apply euler in bold part...5 and 40 are nt co prime

I have not applied Euler in that part...I just calculated the remainder of 7^9 with 16....Then I represented 7^9 as (16K + 7), then I segregated common multiple from numerator and denominator. After that I applied Euler of 8....

What is the O.A.??

Commenting on this post has been disabled.

Krazy me
@Krazy_me
2,294

Ani1308 Sayshi..a little help in remainder part...find the rem of (3^5^7^9)/41

(3^5^7^9)/41

3 and 41 are coprimes.

=> E(41) = 40

5^7^9/40 => 40 = 5*8

=> 5^7^9/8 => E(8 ) = 4

=> 7^9/4 = 3^9/4 => -1 or 3 remainder

=> 5^3/8 = 125/8 => 5 remainder

5^7^9 => 5x = 8y+5 => 5 remainder

3^5/41 = 3*81/41 = 120/41 = 38 remainder

Sorry for x blunder.:-(

No problem.

- 1 Like

Commenting on this post has been disabled.

Ani Rai
@Ani1308
349

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1

=> 5^(16K+7) % 40 = 5

3^(40K + 5) % 41 = (3*81) % 41 = -3 =38

The remainder is38

but dude u cant apply euler in bold part...5 and 40 are nt co prime

Commenting on this post has been disabled.

Varun Tyagi
@varun.tyagi
14,220

Ani1308 Sayshi..a little help in remainder part...find the rem of (3^5^7^9)/41

E(41) = 40

We need to find 5^7^9 % 40

E(40) = 16

7^9%16 = 7

5^(16K+7) % 40 = 5 * 5^(16K+6)%8

Now, E(8 ) = 4

Therefore => (5^16K + 6) % 8 = 1

=> 5^(16K+7) % 40 = 5

3^(40K + 5) % 41 = (3*81) % 41 = -3 =

The remainder is

Commenting on this post has been disabled.

Ani1308 Sayshi..a little help in remainder part...find the rem of (3^5^7^9)/41

C(41)=40

checking for 5^7^9 mod 40

C(40) = 4

7^9 mod 4= 3

so... 5^3 mod 40=5

and 3^5 mod 41 = 38

Commenting on this post has been disabled.

When you follow a discussion, you receive notifications about new posts and comments. You can unfollow a discussion anytime, or turn off notifications for it.

455 people follow this discussion.
0Comments »new comments