@chillfactor said:N = 123456789101112...9899100= 1234567890*10^92 + 101112...9899*1000 + 100For 10111213....9899 remainder by 11 will be 10 + 11 + ... + 98 + 99 or 45*109remainder when N is divided by 11 will be:-6*1 + 45*109*1000 + 1or 6 + 1*(-1)*(-1) + 1or 8So, remainder will be 8
@vbhvgupta said:the square of the sum of the roots of a quadratic equation E is 8 times the product of its roots. Find the value of the square of the sum of the rootsdivided by the product of the roots of the equation whose roots are reciprocals of those of E.
@vbhvgupta said:how many roots (both real and complex) does (x^n - a)^2 = 0 have?
@vbhvgupta said:the square of the sum of the roots of a quadratic equation E is 8 times the product of its roots. Find the value of the square of the sum of the rootsdivided by the product of the roots of the equation whose roots are reciprocals of those of E.
@vbhvgupta said:The lowest possible degree of an equation with real coefficient two of whose roots are sqrt3 and 3+2i is.
@jain4444 said:How many 2007-digit numbers exist, in which every two-digit number composed of two sequential digits is divisible either by 17 or 23?A) 5B) 6C 7D) 9E) More than 9
@ScareCrow28 said:Sir, I couldn't understand the question properly because as for me there will be a hell lot of numbers.2 digit divisible by 17 = 34 and by 23 = 23So there'll be many numbers ? Place 23 or 34 in any 2 positions out of 2007
@jain4444 said:any 2 digit number made by consecutive digits of 2007 digit number will be divisible by 17 or 23 say , 46851 46 mod 23 = 0 68 mod 17 = 085 mod 17 = 0 51 mod 17 = 0
@RDN said:17,34,51,68,8523,46,69,92004692346851792334answer = 9?@jain4444rough work hai upar ka...it appears that all we need to do is select a valid two digit no. for the last two digits of the number, uske baad number apne aap banna shuru ho jaata hai from right to leftthis is bcoz the set of 2 digit multiples of 17 or 23 has a special pattern...each of them have unique last digitsmost interestingly, this means even if we were asked for the number of possible 5 digit nos., or 5009-digit nos., answer would be 9 in each case...
@RDN said:17,34,51,68,8523,46,69,92004692346851792334answer = 9?@jain4444rough work hai upar ka...it appears that all we need to do is select a valid two digit no. for the last two digits of the number, uske baad number apne aap banna shuru ho jaata hai from right to leftthis is bcoz the set of 2 digit multiples of 17 or 23 has a special pattern...each of them have unique last digitsmost interestingly, this means even if we were asked for the number of possible 5 digit nos., or 5009-digit nos., answer would be 9 in each case...
@jain4444 said:How many 2007-digit numbers exist, in which every two-digit number composed of two sequential digits is divisible either by 17 or 23?A) 5B) 6C 7D) 9E) More than 9
@abhishek.2011 said:well there is a doubt number terminates after the last digit apperas as 7in 46923468517 ,after 7 there could not be any digit even after u place 00 still 70 will be a pair which is neither divisibile by 17 nor 23 . how could a 2007 digit number be formed huhh
@abhishek.2011 said:well there is a doubt number terminates after the last digit apperas as 7in 46923468517 ,after 7 there could not be any digit even after u place 00 still 70 will be a pair which is neither divisibile by 17 nor 23 . how could a 2007 digit number be formed huhh
@jain4444 said:How many 2007-digit numbers exist, in which every two-digit number composed of two sequential digits is divisible either by 17 or 23?A) 5B) 6C 7D) 9E) More than 9
a cryptarithmatic problem
@techgeek2050 said:a cryptarithmatic problemSEND + MORE = MONEY how is that? and is there any convention that is generally followed in such questions?
