Question

Two three-digit numbers a and b are such that b has 3 more tens than units in a and 2 units
less than the number of units in a. The sum of a and bis 462 and the difference of aand bis
divisible by 7. What is the sum of the digits of a?​

Answer 1

Step-by-step explanation:

The number lies between 300 and 400.

Thus, we know its a three digit number so let the number be abc.

Now, the value of number will be 100a+10b+c.

On reversing, the number formed will be 100c+10b+a

Sum =100a+10b+c+100c+10b+a=888

=101a+20b+101c=888

Now, the other condition is when units and tens places are interchanged.

So, the number formed will be 100a+10c+b

This number exceeds 100a+10b+c by 9

So 100a+10b+c+9=100a+10c+b

9c−9b=9

c−b=1

c=b+1

Now, as the number lies between 300 and 400, the value of a has to be 3.

So putting value in first equation we get :

101a+20b+101c=888

101∗3+20b+101(b+1)=888

303+20b+101b+101=888

404+121b=888

121b=484

b=4

c=4+1=5

Number will be 345.