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1/X + 1/Y = 1/3....How many ordered pair (x.y) satisfy this.

19 Answers
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Let x be p more than 3 and y be q more than 3
then we have

(1/3+p)+(1/3+q)=1/3
3(6+q+p)=9+pq+3p+3q
18+3q+3p=9+pq+3p+3q
9=pq
now pq can be (1,9),(3,3),(9,1) and x and y can be(4,12),(6,6),(12,4) 

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 For how many ordered pairs (a,b) where a,b are non-negative integers satisfy the following equation:

cube root a + cube root b = cube root of 4160

Hello Apoorva! 

 

x⅓ + y⅓ = 4160⅓ 

 

a• (65)⅓ + b ( 65)⅓ = 4 • (65)⅓ 

 

 

a + b = 4 

( 0, 4) ( 4, 0) ( 1 , 3) ( 3 , 1) ( 2 , 2) 

 

5 ordered pairs 

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what is the value of n if value of n! = 3! *5!*7!

 

Hello Nishant! 

 

3! × 5! × 7! 

 

3! × 5! = 3 × 2 × 5 × 4 × 3 × 2 = 10 × 9 × 8 

 

Hence n ! = 10 × 9 × 8 × 7! = 10! 

IMG 20180703 204509

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New Doc 2018 08 01

Sir please solve this question. 

Hello Samyak , 

 

Is it 8 ? 

No sir, the answer is 7

P² - 4 =( P - 2)( P +2) 

 

Here , P - 2 and P + 2 are primes . [ Difference between primes => P + 2 - ( P -2) = 4 ] 

 

7 and  11 => P = 9

13 and  17 => P = 15

19 and  23 => P = 21

37 and  41 => P = 39

43 and  47 =>P =  45

67 and  73 => P = 69

79 and 83 => P = 81

7 values. 


 

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p is a prime number and m is a positive integer. How many solutions exist for the equation:

p^6 - p=(m^2 + m + 6)(p-1) ?

Hello Manish, 

 

p6 - p = (m2 + m + 6)(p - 1)

 

p( p5 - 1) = (m2 + m + 6)(p - 1) 

 

p ( p4 + p3 + p2 + p + 1 )(p - 1) = (m2 + m + 6)(p - 1) 

 

p ( p4 + p3 + p2 + p + 1 ) = (m2+ m + 6)  

 

If p is a prime greater than 2 then LHS is an odd number , but RHS is even for all integer value of m 

 

So no solution for p> 2.

 

When p = 2 then m2 +  m - 56 = 0 so m= -8, 7 Only one solution m=7 ; p=2 

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Hi sir, this is a question from CopyCat-3

save

My doubt is-- S(n) gives the sum of k and not digital sum, right? Had it been digital sum, then S(555) were equal to 6 and not 15, as stated in the question.

So, 2 digit number n where S(n)=5 can be only 14,23,32,41&50. How is S(59)=5? 
If S(555)=15, then S(59) should be equal to 14 and not 5, right?

Please help.

S((n)) =S(S(n)) now S((59)) will give answer 5 and now solve the question

oh, I did not notice it before. Thankyou!

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shortcut

Sir, is there any easy way to solve this?

I can suggest this

Bdw it is a great question

there are 9 1-digit numbers(9)

there are 90 2-digit numbers(180)
there are 900 3-digit numbers(2700)
and 9000 4-digit numbers(36000)
writing them all gives us 38889 digits.
so the 38890th digit is 1 of the number 10000, the first 5 digit number.
from here on, for the next 10000 numbers (or 50000 digits), each 5th digit would be 1. so 1 is the 38890th digit and 38895th digit etc... and also the 40000th digit.

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How many positive integral pairs of (x, y) satisfy y2 = x4 + x2 + 1?

No such pairs ?

please share your approach

x^4 + x^2 +1 -y^2 =0

Since both roots are +ve c/a >0 

1-y^2>0

1>y^2 

So no possible value of y to be integer 

Hence no such pairs 

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int

120 - y is a perfect sq
y can take 10 different values ( All perfect squares less than 120) 
so, 10 different values for x

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innt

 Denominator ( a - b)^3  is a multiple of 3 so ( a-b) must be a multiple of 3 Option (C). 

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aaaaaaaaaaaaaaaaa

first term : 1 
Sum of first 3 terms : 2 
Sum of first 5 terms : 3 
Sum of first 7 terms : 4 
......................................

Sum of first n terms : 2016 
n = 2 x 2016 -1 = 4031

Sum of the digits : 4 + 0 + 3 + 1 = 8

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1/a+1/b+1/ab=1/n, where a and b are prime numbers and n is a whole number. Find a^2+b^2+n^2

1. 6

2. 10

3. 3

4. None of these

 

I got a and b as 2 and 3, and n=1, so the answer comes 14 which is none of these. But, the given answer is 6. How?

Hello Simran

14 is the correct answer. 

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How many four digit positive integer divisible by 7 have property that, when the first and last digit is interchanged, the result is a (not necessarily four digit) positive integer divisible by 7?

ABCD – DBCA

1000A + 100B + 10C + D – ( 1000D + 100B + 10C + A)

999A – 999D

999( A-D)

999 is not divisible by 7 . so ( A – D) should be a multiple of 7.

(A- D) =   ( 9 – 2) , ( 8 -1 ) , ( 7-0), ( 2 – 9) ,( 1-8) , and ( 1 -1) , ( 2-2) ,….,  ( 9 – 9) total 9 + 5 = 14 cases

Now for each of these 14 cases there are 15 multiples of 7 .
[ 1. 9002 , 9072, 9172, ……., 9982 ( 15 numbers )
  2. 8001 , 8071, ……….., 8981  ( 15 numbers ) similarly others ]

Hence Total : 14  15 = 210 such numbers .

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all the page numbers from a book are added, beginning at page 1. However one page number was added twice by mistake. the sum obtained was 1000. Which page number was added twice?

n*(n+1)/2 = approx less than 1000
n(n+1) <2000

square root of less than 2000 is 44^2 = 1936
so again , if all the numbers till 44 are added then the sum is following
44*45/2 = 990
Hence 10 was added twice

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