Question

There is a certain number consisting of 3 digits which is equal to 13 times the sum of the digits and if 495 is added to the number, the digits will be reversed. Also the sum of the extreme digits is greater than the middle digit by 2. Find the number.

Answer 1

Answer:

Let a 3-digit number be abc

this implies abc = 13(a+b+c)

100a+10b+c=13a+13b+13c

87a-3b-12c=0(÷by3both side)

29a-4c=b ---- eq(1)

Also a+c =b+2------eq(2)

a+c=29a-4c+2

a-29a+c+4c =2

-28a +5c =2 -------eq(3)

on reversing the digits we have

100c+10b+c=100a+10b+a+495

100c-c+a-99a=495

99c-99a=495(÷both sides by99)

we get c-a=5------eq(4)×by5

5c-5a=25----eq(6)

on subtracting eq(3) from eq(6)

we get 23a =23

therefore a=1

substitute a=1 in eq(4) we get

c=6

now put c=6 in eq(2)

we get b=5

therefore abc=156