Question 80

# Virat was asked to multiply a two-digit number P by a three-digit number Q. But she mistakenly multiplied P by the number formed by writing the digits of Q in the reverse order, there by getting an answer which is 22770 more than the correct answer. What is the minimum possible sum of the digits of Q?

Solution

Let's assume the two digit number P as uv=u*10+v

Let's assume the 3 digit number Q as xyz= x*100+y*10+z

The actual multiplication result is uv*xyz=(u*10+v)*( x*100+y*10+z)

But he reversed the digits of Q. So, Q is considered as zyx = z*100+y*10+x

The result of multiplication after this operation is uv*zyx=(u*10+v)*( z*100+y*10+x)

Given the difference between these is 22770.

(u*10+v)*( z*100+y*10+x) - (u*10+v)*( x*100+y*10+z)=22770

Let's assume (u*10+v) as t.
t*( z*100+y*10+x) - t*( x*100+y*10+z)=22770

100zt+10yt+xt-100xt-10yt-zt=22770

99zt-99xt=22770

99(zt-xt)=22770

zt-xt= 230

(z-x)*t=230

z and x are two single digit numbers. So, the difference will also be a single digit number. And t is as two digit number.

Let's check the possible values of these.

230 can be written as 2*115, 5*46, 10*23

2*115: this is not possible because t is not a 3 digit number.

5*46 is possible.

10*23 is not possible because z-x is not a two digit number.

So, the only possible value is z-x=5 and t=46

The question ask the minimum value for sum of digits of Q. The minimum value we can give for z,x is 6,1. And the minimum value of y is 0.

So, the sum is 6+1+0=7