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Today we will discuss a problem, I am not giving it as an exercise as I want to discuss it, rather than discuss a 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

d=4 (1,1,4) (1,2,2) (7*3*4),(4*4*4),(8*8*1)(9*8*2) and others

d=4 works

d=5 is obvious (5*1*3)

d=6 is (1*1*6) ,(1*2*3) (4*4*1)

which works

hence d=6 works too

d=7  works for 7*1*1

d=8 (4*2*1) (6*8*1) (6*2*4)(2*2*2) (2*3*8)

d=8 does not work

d=9 will work for (9*7*3)

so d=3,8 does not work

Hence 2 digits cant be unit digits !


Written by Implex

September 22, 2008 at 9:24 pm

Posted in Number Thoery

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