Number question

Question

if z=a*b*c where a,b, c are consecutive integers in that order, and a is an even number. Which of the following must also be an integer. 

1. z/3

2.z/4

3.z/8 

a. 1 only 

b. 2 only 

c. 1 and 2 only 

e. 1,2 and 3 

Answer 1

Consider z=a*b*c.

 

  Multiples of 3 occur every third integer, so one of (a, b, c) must be a multiple of 3.  That means that (z/3) is an integer.   Ans [1] is true.

  If a is even, we could represent that by 2k.  Then,
      z = (2*k)*b*c
      (z/2) = k*b*c      [that is, z must be evenly divisible by 2]
   and
     if a, b, and c are consecutive integers, and a is even, then c must be even [c=2j], and
      (z/4) = k*b*j    [where k and j are integers]

 So, [2] is true.

Is [3] always true?

   For these values of a, b, and c, (z/8) is not an integer:

       1   2   3
       3   4   5
       5   6   7

       9 10 11
     11 12 13

     13 14 15

 

     17 18 19
     19 20 21
     21 22 23

 

     25 26 27

      . . .      

No, Ans [3] is not always true.

 

So, Ans [c]

Answer 2

z=a*b*c where a,b,c are consecutive integers

one of every three consecutive integers is a multiple of 3, no matter where you start

1,2,3

2,3,4

3,4,5

4,5,6

5,6,7

etc

 

a. is the answer because a*b*c is a multiple of 3 and therefore divisible by 3