Question

Three digit numbers such that its digit at tens place is one more than the digit at hundreds place the sum of the digits is 13 the number obtained by reversing the digits can be obtained by multiplying the original number with two and subtracting 49 from this product find the original number

Answer 1

so the original number is 346. .

Answer 2

Answer:346

Step-by-step explanation:

let the number is 100x+10y+z here x,y,z represents hundreds, tens and unit places. i use this number so that we have to calculate less digits.

as given

x=y-1

sum of digit is = 13

x+y+z =13

⇒y-1+y+z=13 (x=y+1)

⇒2y+z=14 ----------- 1

and

100z+10y+x = 2( 100x+10y+z)-49

⇒100z+10y+x=200x+20y+2z-49

⇒98z-10y-198x=49

⇒98z-10y-198(y+1)=49

⇒98z-10y-198y-198=49

⇒98z-208y=247 --------2

by solving equation 1 and 2

y=4 and z= 6

by putting y value in (x=y+1)

we get x=3

so number formed is 346