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 !
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