For how many integers from 10 to 30 is the remainder when [m

Question

For how many integers  x  from 10 to 30 is the remainder when  x^2 - 3x  is divided by 30 equal to 4?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89324 --- Harsha GMAT-Club-Forum-pw3kx7hm.png

TALHA23 · Answer

hi can someone please solve this question for me using very time effiecient approach im having trouble solving it even tho it isnt a complicated question but im not able to wrap my head around it . thank u

RaahiK · Answer

We want x^2−3x−4, so it's exactly divisible by 30.

=> x^2−3x−4 =(x−4)(x+1) AND we know 30=2×3×5 (the product must be divisible by 2, 3, and 5)

Consider 5:  either x−4 or x+1 is divisible by 5
If x−4 is divisible by 5, then x=4, 9, 14, 19, 24, 29... AND if x+1 is divisible by 5, then x=4, 9, 14, 19, 24, 29... so it's the same list.

Between 10 and 30, the possibilities are  14, 19, 24, 29

Consider 3:
 14: (14−4)(14+1)=10×15=150 divisible by 30
19: 15×20=300 divisible by 30
24: 20×25=500  not  divisible by 30
29: 25×30=750 divisible by 30

So we have 3 values: 14, 19, 29.

Answer = 3

kaustubhkohli12 · Answer

Why between 10 and 30, because we are looking for multiples of 5 ?

kevincan · Answer

If the remainder when  x^2 - 3x  is divided by 30 is 4 ,  x^2 - 3x -4  is a multiple of 30.
 x^2 - 3x -4= (x-4)(x+1)  is thus a multiple of 2,3 and 5

Note the difference between  (x+ 1)  and   (x -4)  is 5. This tells us that these terms have opposite parities (one is odd and the other is even) and that either both are multiples of 5 or neither is multiple of 5. The product of these two terms will be even regardless of the value of x.

We can therefore say that the product will be a multiple of 30 if and only if x-4 is a multiple of 5, and either X -4 or x + 1 is a multiple of 3.

For x-4 to be a multiple of 5, x must be either 14,19,24 or 29.

14+1 is a multiple of 3, 19-4 is a multiple of 3, 29+1 is a multiple of 3. Neither 24-4 nor 24+1 is a multiple of 3.

Thus the remainder is 4 for three values of x : 14,19 and 29

HarshavardhanR · Answer

Hey!

Observe that we got (x-4)(x+1) = some multiple of 30.

In other words,

(x-4)(x+1) = (2 x 3 x 5) x (something).

Observe that (x-4)(x+1) contains a 5.

So, at least one of the terms x-4 or x+1 has to contain a 5 (else, where will the 5 come from?).

The above means that  we only need to worry about cases where at least one of the other terms is a multiple of 5 . Hope this answers your core question!

----

In addition, here is another fun observation.

If (x-4) is a multiple of 5, then x+1 is also a multiple of 5. This is because x+1 = (x - 4) + 5 = 5k + 5 = 5a.
Similarly, if (x+1) is a 5-multiple, so is x-4. x - 4 = x+1 - 5 = 5k - 5 = 5b.

In other words, it so happens that if one of the two numbers is a 5-multiple, so is the other.

Hence, we could just go one by one, checking only those x values for which (x-4) values are 5-multiples. This will, by default, also include every (x+1) values which are 5-multiples.

Hope this helps!
Harsha

egmat · Answer

Hi kaustubhkohli12,

Good instinct to ask where a number comes from - but in this case the  10  to  30  has nothing to do with the 5s. It comes straight from the question stem.

Read the stem again:  "For how many integers x from 10 to 30..."  That range is the question's own restriction on x - the pool of candidates you are allowed to count, handed to you before you do any work at all. If the stem had said "from 1 to 100," the entire solution would run the same way, just on a bigger pool.

The two filters are doing different jobs

The 5s do something else entirely. Since x^2 - 3x - 4 factors as (x - 4)(x + 1), and  30  =  2  ×  3  ×  5 , one of those two factors has to carry the  5 . If x - 4 is a multiple of  5 , then x =  4, 9, 14, 19, 24, 29 , ...; if x + 1 is a multiple of  5 , you land on that same list. Either way the condition reduces to one thing:  x is 4 more than a multiple of 5 . And that list runs on forever.

Now the stem's range does its job: keep only the entries that sit between  10  and  30 , and you are left with  14, 19, 24, 29 . That is where those four candidates came from - the 5-condition supplies the pattern, the stem supplies the window.

What is left to check

Two factors of  30  remain. The  2  is free: x - 4 and x + 1 always have opposite parity, so their product is always even. The  3  is not free: x must not be a multiple of  3 , which knocks out  24 .

That leaves  14 ,  19 ,  29  -  three  values.

Answer: B

For how many integers from 10 to 30 is the remainder when [m

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For how many integers from 10 to 30 is the remainder when [m

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