Problema Solution
there is a certain number consisting 4 digits which is equal to 202 times the sum of the digits.The sum of original number and the number obtained by interchanging the digits in thousand's place with the digit in hundred's place and the digit in ten's place with the digit in unit's place is 3333.Also the digits in thousand's and ten's place is same as well as the digit in hundred's place and unit's place are equal.Find the original number. [Ans:1212]
Answer provided by our tutors
let
x = the digit in thousand's and ten's place
y = the digit in hundred's place and unit's place
the number can be written as 1000x + 100y + 10x + y
1000x + 100y + 10x + y = 202(x + y + x + y)
1000x + 100y + 10x + y = 202*2(x + y)
606x - 303y = 0 divide both sides by 303
2x - y = 0
the sum of original number and the number obtained by interchanging the digits in thousand's place with the digit in hundred's place and the digit in ten's place with the digit in unit's place is 3333
1000x + 100y + 10x + y + 1000y + 100x + 10y + x = 3333
1111x + 1111y = 3333 divide both sides by 1111
x + y = 3
by solving the system of equations
2x - y = 0
x + y = 3
we find
x = 1
y = 2
click here to see the step by step solution of the system of equations
the original number is 1212.