The product of 3 consecutive positive multiples of 4 must be divisible by each of the following EXCEPT
a.16
b.36
c.48
d.128
e.192
Does anybody know a fast way of solving this question? Thanks.
multiples Q.
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IMO B
since there are three consecutive positive multiples of 4
4K*4(k+1)*4(k+2)
A out
C48=16*3, we already have 4*4=16. and of any three consecutive integers, one must be a multiple of 3. out
D128=4*4*4*2. we already have 4*4*4. and of three consecutive integers at least one must be a multiple of 2. out
E. 192= 4*4*4*3. same reasoning as C out
B36=4*3*3. we may or may not get two 3's
answer B
since there are three consecutive positive multiples of 4
4K*4(k+1)*4(k+2)
A out
C48=16*3, we already have 4*4=16. and of any three consecutive integers, one must be a multiple of 3. out
D128=4*4*4*2. we already have 4*4*4. and of three consecutive integers at least one must be a multiple of 2. out
E. 192= 4*4*4*3. same reasoning as C out
B36=4*3*3. we may or may not get two 3's
answer B
The powers of two are bloody impolite!!