If X,Y and Z are three consecutive multiples of 5. Then which of the

Question

If X,Y and Z are three consecutive multiples of 5. Then which of the following will the factor of X*Y*Z
1) 125
2) 250
3) 375
4) 750

A) only 1
B) 1,2
C) 1,3
D) 1,2,3
E) All of them

*** Just out of curiosity can X+Y+Z be divisible by 15 or 30 ?

Argo · Answer

Given choices can be re-written as multiples of 125

1. 125
2. 125*2
3. 125*3
4. 125*6

Now if, 5*a = X, then

5*(a+1) = Y & 5*(a+2) = Z

So, X*Y*Z = 125* a(a+1)(a+2)

Now, among any 3 consecutive positive integers, we will either have (a number that is divisible by both 2 & 3) or (a number divisible by 2 and another number divisible by 3).

Ex: 
17, 18, 19 -> 18 divisible by 2 & 3.
14, 15,16 -> 14, 16 divisible by 2 & 15 divisible by 3.

Hence 'E'

preetamsaha · Answer

If X,Y and Z are three consecutive multiples of 5. Then which of the following will the factor of X*Y*Z
1) 125
2) 250
3) 375
4) 750

let,X=5a
Y=5(a+1)
Z=5(a+2)
so,X*Y*Z=5^3(a)(a+1)(a+2) = 125(a)(a+1)(a+2)
now,product of 3 consecutive positive integers is always divisible by 6.
so,X*Y*Z is always divisible by 
1.125
2.125*2= 250
3.125*3= 375
4.125*6= 750
 
correct answer E

NOTE : 
Let three consecutive positive integers be, n, n + 1 and n + 2.
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
 
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
 
Since, n (n + 1) (n + 2) is divisible by 2 and 3.
 
∴  n (n + 1) (n + 2) is divisible by 6.

TestPrepUnlimited · Answer

Since they are consecutive multiples of 5, each letter must have a multiple of 5. If we divide all letters by 5 we would get three consecutive integers, therefore X*Y*Z must also contain a factor of 3 and a factor of 2.

(1) 125 = 5 * 5 * 5 . We can take one 5 out of each letter so there must be a factor of 125.
 (2) 250 = 125 * 2 . We have 125, and there must be a multiple of 2 as well since there are 3 consecutive multiples
 (2) 375 = 125 * 3 . We have 125, and there must be a multiple of 3 as well since there are 3 consecutive multiples.
 (3) 750 = 6 * 125.  Same as above, it must be contained in X*Y*Z.

Therefore all of them must be a factor of X*Y*Z.

Ans: E

And as a note X + Y + Z must have a factor of 5, and again they are multiples of some consecutive sequence of 3 integers, so their sum must also be a multiple of 3. We cannot conclude anything else so we can only say X + Y + Z is divisible by 15, but not necessarily 30 since 30 = 6 * 5, and we cannot confirm if the sum is even.

tam87 · Answer

Taking 5x1, 5x2, and 5x3 as the values of x, y, z
x * y * z = 125 * 6 = 750

As it is divisible by 750, it will also be divisible by any factor of 750
All the options are factors of 750

Hence, option E

Add: x+y+z should be a multiple of 15 always, but not necessarily a multiple of 30

Posted from my mobile device

Abhishek009 · Answer

5*1 = 5
5*2 = 10
5*3 = 15

So, a*b*c = 5*10*15 => 750

1. 750/125 = 6
2. 750/250 = 3
3. 750/375 = 2
4. 750/750 = 1

Hence, Answer must be all of them (E)

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