[MaskedMenace]

Originally Posted by shilpakokkanti

If the hands of a clock coincide every 80 minutes and the hands of another clock coincide evry 65 minutes..What is the app time difference between the two clocks in exactly 24hrs time interval as shown by a correct clock?

80 mins = 1hour

In another clock, 65 mins = 1 hour.

in 24hours, 24*(80 - 65) = 360 mins

as shown by correct clock, it is 6hours.

[prakharc]

Originally Posted by $U!\!$H|!\!E

What would be the total number of factors of the expression
62^3 -54^3-8^3?

62^3 -54^3-8^3=
54(62^2 + 8^2 +496) - 54^3 = 54*4[31^2 + 4^2 + 124-27^2]
= 54*4[4*58 + 4^2 +124] = 4^2*54*3*31=2^3 * 3^4 * 31

Total numebr of factors= 4*5*2= 40 (including 1 and number itself)

[MaskedMenace]

<snip>wrongpost

[MaskedMenace]

Originally Posted by $U!\!$H|!\!E

What would be the total number of factors of the expression
62^3 -54^3-8^3?

(54+8 )^3 - 54^3 - 8^3

54^3 + 8^3 + 3*54*8(62) - 54^3 - 8^3

3*54*8*62

3*2*3^3*2^3*2*31

31*2^5*3^4

2*6*5 = 60 factors

[$U!\!$H|!\!E]

Originally Posted by prakharc

62^3 -54^3-8^3=
54(62^2 + 8^2 +496) - 54^3 = 54*4[31^2 + 4^2 + 124-27^2]
= 54*4[4*58 + 4^2 +124] = 4^2*54*3*31=2^3 * 3^4 * 31

Total numebr of factors= 4*5*2= 40 (including 1 and number itself)

Actually answer is 60
54*4(4*58+4^2+124)
=> (2*3^3)(2^2)*4(58+4+31)
=> (2)^5*3^3*93
=> (2)^5*3^4*31

So no of factors: (5+1)*(4+1)*(1+1) = 60

Thanks

[$U!\!$H|!\!E]

Originally Posted by MaskedMenace

(54+8 )^3 - 54^3 - 8^3

54^3 + 8^3 + 3*54*8(62) - 54^3 - 8^3

3*54*8*62

3*2*3^3*2^3*2*31

31*2^5*3^4

2*6*5 = 60 factors

Thank you!

[Tripti11]

please tell me the solution using the remainder theorem..

what is the remaindewr when x^100-2(x^99)+x^50-2(x^49)+x+1 is divided by (x-1)(x-2) ..?

[$U!\!$H|!\!E]

Shakuntala asked Aryabhatta to assume any two values of three digits say P and Q. Then she told him to multiply P by R and Q by S where the values of R and S were given by Shakuntala herself.

Aryabhatta exactly told her the value of PR+QS = 888222.

Then Shakuntala told him the value of P+Q. Can you find out that value?

[shanks4mba]

Originally Posted by $U!\!$H|!\!E

Shakuntala asked Aryabhatta to assume any two values of three digits say P and Q. Then she told him to multiply P by R and Q by S where the values of R and S were given by Shakuntala herself.

Aryabhatta exactly told her the value of PR+QS = 888222.

Then Shakuntala told him the value of P+Q. Can you find out that value?

its 8 + 2 = 10
approach...
here R and S should be 1 and 1000 to knw both the nos.
hence we get P = 2 and Q = 8
hence P + Q = 10

[$U!\!$H|!\!E]

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