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

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)

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

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!

@nishant thank you for this , i have also some confusion that's why i search here.

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.

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,

p^{6} - p = (m^{2} + m + 6)(p - 1)

p( p^{5} - 1) = (m^{2} + m + 6)(p - 1)

p ( p^{4} + p^{3} + p^{2} + p + 1 )(p - 1) = (m^{2} + m + 6)(p - 1)

p ( p^{4} + p^{3} + p^{2} + p + 1 ) = (m^{2}+ 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 m^{2} + m - 56 = 0 so m= -8, 7 Only one solution m=7 ; p=2

Hi sir, this is a question from CopyCat-3

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!

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.

How many positive integral pairs of (x, y) satisfy y^{2} = x^{4} + x^{2} + 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

120 - y is a perfect sq

y can take 10 different values ( All perfect squares less than 120)

so, 10 different values for x

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

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

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.

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 .

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