Approach: Numbers having 1/0 in them will not satisfya! * b! > a! + b! as 1 * something is always less than 1 + something.So eliminate numbers having 1/0 in them11-91, 10-90, 12-19= 26 numbersAlso, an exception is 22 where 2! + 2! = 2! * 2!Hence, total 27 numbers will not satisfy.
So, 90-27=63 numbers will satisfy
[solved by vinit sanghvi]
Q30) P and Q run on a circular track of 100m. Speed of P is 250 m/s and that of Q is 400 m/s. In how many points on a circular track will they meet if they start simultaneously from the same point, in the same direction?
If N = p1^a + p2^b + p3^cthen ordered factor pairs coprime to each other = (2a + 1)(2b + 1)(2c + 1)Unordered factor pairs coprime to each other = {[(2a + 1)(2b + 1)(2c + 1) - 1]/2} + 1 (for (1,1)
360 = 2^3 * 3^2 * 5Ordered factor pairs coprime to each other = (2 * 3 + 1) * (2 * 2 + 1) (2 * 1 + 1) = 7 * 5 * 3 = 105Un ordered factor pairs coprime to each other = (105 - 1)/2 = 52 + 1 = 53if we don't consider (1,1) then 52.Confused!
Detailed approach (Hemanth Venkatesh)
On factorizing 360 we get the following factors:360 = (2^3) * (3^2) * (5)First suppose that 5 divides neither co prime factor. Then we have two cases:Case 1: One factor must be 2 to some positive power while the other factor must be 3 to some positive power. Hence there are 3 * 2 = 6 of these (as there are 3 possible positive powers of 2 and there are 2 possible positive powers of 3).Case 2: One factor is 1 while the other factor is a divisor of (2^3) * (3^2).Hence there are (3+1)(2+1)=12 of these.
Now, note that all of these pairs consist of different numbers except the pair {1,1}. For each of the pairs with two distinct elements, we can multiply either element by 5 to get two new co prime pairs of factors (that is the solution {a,b} gives rise to the solutions {5a,b} and {a,5b}.Finally, the pair {1,1} under this process only gives us one new pair: {1,5}.In conclusion, we have 3(6 + 12 − 1) + 2 = 53 co prime pairs.
General term - nCr * (x^2)^(n-r) (3y)^r
It is given that 2n - 2r + r = 52n - r = 5n = 4 => r = 3
So the term we have here is 4C3 * (x^2)^1 (3y)^3Coefficient = 4C3 * 3^3 = 4 * 27 = 108
@Rowdy-Rathore
F(x) = x^4-360x^2+400F(x) = (x^2+20)^2 - 400x^2F(x) = (x^2-20x+20)(x^2+20x+20) = Prime NumberSince the value of second factor is always positive and greater than 1 therefore, 1st factor hasto be equals to 1.x^2-20x+20 = 1x^2-20x+19= 0On solving,x = 1, 19.Putting back the values in the original equation we get f(x) = 41 and f(x) = 761.
So, the sum will be 41+761 = 802
