Reply To: Challenge Questions – 2

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Satya
Participant

Challenge question 15:

321 x ? ? ? = ? ? ? 123

I used vertically and crosswise multiplication method to find the missing digits.

From the above sentence it is clear that the ending digit of the second multiplier is 3

3 2 1

x y 3

———————-

a b c 1 2 3

(2*3) + (1*y) should be equal to a 2 digit number that ends in 2 i.e. 12 or 22 or 32 etc. y here has to be single digit so y has to be 6.

3 2 1

x 6 3

—————

a b c 0(+1) 2 3

Now (3*3) + (2*6) + (1*x) = multiple of 10 –> 21 + x = multiple of 10. Since x has to be single digit, it should be 9

3 2 1

9 6 3

————-

a b c 1 2 3

=  3 0 9 1 2 3

  • This reply was modified 7 months, 2 weeks ago by Satya.
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