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Improve your reasoning power and Solve the Mathematics equation problem

Improve your reasoning power Question - When a no. is divided by 2, rem. is 1 when divided by 3, rem. is 2. and so on...till it is divided by 10, remainder is 9.What's the number? Answer- if the no. be n, (n+1) is divisible by all of {2,3,4,..,10}. hence (n+1) is divisible by 2520. [ 2 |(n+1) & 3 |(n+1) => 6 |(n+1) & 4 | (n+1) => 12 |(n+1) (12=l.c.m(6,4)) & 5 |(n+1) => 60 |(n+1). continuing in this way 2520 |(n+1)] therefore any no. of the form n=(2520p-1) will suffice where p is a natural no. of course the smallest such natural no. is 2519. Mathematics equation problem X^3-Y^2=2 FIND ALL INTEGRAL SOLUTION OF X,Y Answer: case 1 x-3=y-5 y=x+2 original eq gives x^3-x^2-4x-2=0 no integral solution for this. case 2 x-3=y+5 y=x-8 original eq gives x^3-x^2+16x-66=0 (x-3)(x^2+2x+22)=0 x=3 y=-5 as eq invoves y^2 hence y as wellas -y both satisfies the pairs are (3,5) and (3,-5)

Find the remainder

Find the remainder when the digits 1 to 99 written side by side is divided by 11 i.e: (123456789101112...99)mod 11=? 123456789101112.....979899 counting from right, sum of digits at odd place = 9(9+8+...+1+0)+(9+7+5+3+1) sum of digits at even place = 10(9+8+...+1)+(8+6+4+2) sum of odd - sum of even = -40 = 4 (mod 11)