if n is an odd multiple of 3 how many ways can 2^n be expressed as product of 3 factors
When 2 of the numbers is equal, then its like we're lookin for solns to 2a + b = n2a = n - bFor 'a' to be an integer, since n is odd, b needs to be odd as well.So b can take all odd values from 1 to n. But we exclude the particular value when b = a ( = n/3), since we've already considered the case of all 3 being equal.So total no. of such (a,b) pairs = (n-1)/2
If x and y are even integers,then thenumber of solutions of x^2-y^2 = 7389746232
Let x = 2m and y = 2nThen4(m^2 - n^2) = 7389746232m^2 - n^2 = 1847436558 = 2(923718279)(m-n)(m+n) = 2(923718279)The above expression means that 1847436558 can be expressed as a product of 2 factors, (m-n) and (m+n)Sum of factors = 2m -> even.This is however not possible since 1847436558 = 2(923718279) has only 1 even factor, which is 2. And even + odd = odd.Hence no solutions
How many natural numbers less than 100 when squared and then divided by 24 leave a remainder of 1?
All right, I'll give various ways to solve this problem, take your pick.Method 1.All primes (greater than 3) when squared are divisible by 24.All primes can be expressed as 6k+/-1 -> k can vary from 1 to 16. As also the integer 1 satsifies.So total 33 values.Method 2.We need x^2Mod24 = 1Let x = (12k + n) -> n can take values from 0 to 11x^2 = 144k^2 + 24k + n^2All terms on RHS are divisible by 24 except n^2.So x^2Mod24 = n^2Mod24Values of n that satsify our condition are - 1,3,5,7Total 12k+1 - (k varies from 0 - 8) - 9 values12k+3 - (k varies from 0 - 8) - 9 values12k+5 - (k varies from 0 - 7) - 8 values12k+7 - (k varies from 0 - 7) - 8 valuesTotal = 34 (but remove the number 3 of the form 12k + 3 for k = 0)So total = 33Method 324 = 8 * 3Charmichael number for both 3 and 8 is 2.What that means is that if k is coprime to 24, then k^2Mod24 = 1So all no.s less than 100 coprime to 24 satisfy.Uptill 96, we hav 96(1-1/2)(1-1/3) = 32 such numbers.From 97 - 100, 97 also satisfies.So total 33 numbers.
Remainder when 5353535353......(100 times) is divided by 102 ?
You can read up CRT here:
What would be remainder when N is divided by 23, where N=1313131313......... a 800 digits number
E = 2213 repeated 22 times is divisible by 2313 repeated 22(18) = 396 times is divisible by 23.
A number 'k' repeated E[n] times where n is the divisor is perfectly divisible by the divisor.E[n] is the Euler no. for 'n'.The only condition is that n and k should be coprime, morover, n should be coprime to 2,3 and 5So for eg, if n = 7, E[n] = 6So 555555Mod 7 = 0. ie, 5 repeated 6 times divisible by 7
@giyan Say a number is abcdThis can be represented as 1000a + 100b + 10c + dNow, (1000a + 100b + 10c + d)Mod9 = (a + b + c + d)Mod9 => A number that comes down to a single digit between 1-9So finding remainder when dividing a number by 9 gives you the sum of its digits, down to a single digit.
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