A second way which I like, though longer:If the prime is n, then n^2 - 1 = (n+1)(n-1)Now n is a prime > 2 so is odd. So n+1 and n-1 are even, and that too consecutive even numbers so one of them will also be divisible by 4. Hence the product will be divisible by 2*4.Also one out of any 3 consecutive numbers iwll be divisible by 3. Now in n-1, n, n+1,the prime number n cannot be the one so one of the others must have a factor of 3 as well.So overall (n+1)(n-1) = n^2 - 1 is of the form 24k. So n^2 = 24k+1 and will hence leave a remainder 1 when divided by 24.regardsscrabbler
Excellent work @scrabbler
@saurav205..yes, there is no problem with approach :)
Next now: Find the area of the triangle with sides rt(13), rt(29) and rt(34).
A second way which I like, though longer:If the prime is n, then n^2 - 1 = (n+1)(n-1)Now n is a prime > 2 so is odd. So n+1 and n-1 are even, and that too consecutive even numbers so one of them will also be divisible by 4. Hence the product will be divisible by 2*4.Also one out of any 3 consecutive numbers iwll be divisible by 3. Now in n-1, n, n+1,the prime number n cannot be the one so one of the others must have a factor of 3 as well.So overall (n+1)(n-1) = n^2 - 1 is of the form 24k. So n^2 = 24k+1 and will hence leave a remainder 1 when divided by 24.regardsscrabbler
Excellent work @scrabbler@saurav205..yes, there is no problem with approach Next now: Find the area of the triangle with sides rt(13), rt(29) and rt(34).Team BV - Kamal Lohia
doin it by hero's formula seems a tedious task..!!
1> how many perfect squares are the divisors of the product 1!.2!.3!....8!?
1!.2!.3!....8! can be written, in prime factorised form, as 2^23 * 3^9 * 5^4 * 7^2
Now for a perfect square, each factor should be chosen in pairs and we can choose a power of 2 AND of 3 AND of 5 AND of 7.
Power of 2 can be chosen in 12 ways (2^0, 2^2, 2^4, 2^6...2^22) similarly Power of 3 can be chosen in 5 ways (3^0, 3^2...3^8) Power of 5 can be chosen in 3 ways (5^0, 5^2, 5^4) and Power of 7 can be chosen in 2 ways (7^0, 7^2)
So overall since it is an AND case, fundamental principle of counting tells us that we have 12 * 5 * 3 * 2 = 360 such factors. @mohitjain
Bro you wil have total number of three's as 9 Total number of 5's as 5 Total number of 7's as 2 So the total number of square numbers as divisors will be all the even powers of all the priime factors
2 has 2,4,6,8,....22=11 3 has 2,4,..8=4 5 has =2 7 has =1 So total number of divisors = (11+1)(4+1)(2+1)(1+1)=12*5*3*2=360
@aditi88 refer to my solution Bro you wil have total number of three's as 9 Total number of 5's as 5Total number of 7's as 2So the total number of square numbers as divisors will be all the even powers of all the priime factors2 has 2,4,6,8,....22=11 3 has 2,4,..8=45 has =27 has =1So total number of divisors = (11+1)(4+1)(2+1)(1+1)=12*5*3*2=360
