Software Applications Programming Errors and solutions.

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 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)

Find the missing number in the series? 87431 ---> 815 ---> 23 ---> ? Ans: The number is 5. In this series group the odd and even numbers separately and add them you’ll get the next series of numbers. Similarly perform this till the last number of the series.. ie... 87431=>84+731=815 815=>8+15=23 23=>2+3=5