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Official Quant Thread for CAT 2011 [Part 8]

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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.

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
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Ani1308 Says
hi..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 Says
The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?

The sum of digits of a natural number (10^n 1) is 4707, where n is a natural number. The value of n is ?
10^n-1 is obviously multiple of 9 and 11
so 4707/9 =523 (sum will be 9*n where n is no. by which 9 is repeated)
so value is 9999999999999999............523 times

P.S.:- jain sahab itna easy kaise de sakte hai
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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 1.
E(11) = 10
25^anything has unit digit 5
5^(10K+5) % 11 = 5^5 % 11 = 25*25*5 = 45 = 1
Smoothest seas do not make tough sailors
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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...

Have fun in whatever you do, otherwise whats the point !!
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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 is 38

Ani1308 Says
but 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.??
Smoothest seas do not make tough sailors
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Ani1308 Says
hi..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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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 is 38

but dude u cant apply euler in bold part...5 and 40 are nt co prime
Have fun in whatever you do, otherwise whats the point !!
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Ani1308 Says
hi..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 = 38

The remainder is 38
Smoothest seas do not make tough sailors
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Ani1308 Says
hi..a little help in remainder part...find the rem of (3^5^7^9)/41

i am getting 38 ... oa?

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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