In this lesson I am going to discuss a concepts
straight away by solving a question rather than discussing the
whole new concept.

**How many digits cannot be
the unit's digit of the product of three 3-digit numbers whose sum
is 989**

When we read the problem, it looks
like what the hell is it asking? But, it is not that difficult a
problem if we go by a method

let the three 3 digit numbers be
xyz, abc and pqr

then what the question essentially
says is z+c+r=9 or 19

now we have to check if d is the
last digit of the product of xyz, abc and pqr, then what are the
values d cannot take.

it can be easily checked that it
works for 0 and 2 so d can be o or 2

for d=1, (z,c,r)=(9,3,3) or(9,9,1)
and others

9+9+1=19 satisfies hence d can be
1

d=3, (z,c,r)=(3,1,1),(3,7,3),(9,7,1)
but none of which sum to 9 or 19

hence d cant be 3

and
others

d=4 works

d=5 is obvious

d=6 is

which works

hence d=6 works too

d=7 works for

d=8 does not work

d=9 will work for

so d=3,8 does not work

Hence 2 digits cant be unit digits
!

## Post Comments