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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
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)
9+9+1=19 satisfies hence d can be
but none of which sum to 9 or 19
hence d cant be 3
d=5 is obvious
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 !