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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...

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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

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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

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Varun Tyagi
@varun.tyagi
14,221

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

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Ani Rai
@Ani1308
349

5^25^125^3125 /11..

thanks all...

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Varun Tyagi
@varun.tyagi
14,221

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.??

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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

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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

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Varun Tyagi
@varun.tyagi
14,221

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

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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

2012 :Joined NMIMS. Left after an year due to unfortunate circumstances. Target MBA 2016-18 :D

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