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The sum of all integers between 500 and 2500 that are divisible by bot

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Joined: 12 Oct 2010
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The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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28 Mar 2019, 06:50
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Difficulty:

55% (hard)

Question Stats:

50% (01:55) correct 50% (02:38) wrong based on 30 sessions

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GMATH practice exercise (Quant Class 16)

The sum of all integers between 500 and 2500 that are divisible by both 18 and 75 is:

(A) 6750
(B) 6300
(C) 5400
(D) 4050
(E) 3150

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Joined: 02 Aug 2009
Posts: 7981
Re: The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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28 Mar 2019, 07:50
2
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

The sum of all integers between 500 and 2500 that are divisible by both 18 and 75 is:

(A) 6750
(B) 6300
(C) 5400
(D) 4050
(E) 3150

Let us find the first and the greatest integer of that kind in the given range..

For this, let us find LCM(18,75)=450..
so first integer is 450*2=900..
now 900*3=2700, and 2500 is just less than 2700, so the greatest integer in the range is 900*3-450 = 450*2*3-450=450(6-1)=450*5..

so the SUM will be 450*2+450*3+450*4+450*5 =450(2+3+4+5)=450*14=6300

B
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Re: The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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28 Mar 2019, 08:18
I tried using the sum of series. Since the question asks to find the multiples of both 18 and 75, The multiples of 18 will be a subset of multiples of 75.

For 500 to 2500

a1 = 525
an = 2475
d = 75
n = ((an-a1)/d) +1 = 27

I am lost how to proceed further, and I have exhausted 2 minutes.
Math Expert
Joined: 02 Aug 2009
Posts: 7981
Re: The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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28 Mar 2019, 09:33
anbuKp wrote:
I tried using the sum of series. Since the question asks to find the multiples of both 18 and 75, The multiples of 18 will be a subset of multiples of 75.

For 500 to 2500

a1 = 525
an = 2475
d = 75
n = ((an-a1)/d) +1 = 27

I am lost how to proceed further, and I have exhausted 2 minutes.

The multiples of 18 will NOT be subset of multiples of 75.. 75 is multiple of just 3, while 18 is a multiple of 9..
So, to find first multiple of 18 and 75, we have to find the LCM of 18 and 75.
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Re: The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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28 Mar 2019, 12:05
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

The sum of all integers between 500 and 2500 that are divisible by both 18 and 75 is:

(A) 6750
(B) 6300
(C) 5400
(D) 4050
(E) 3150

$$? = \sum N \,\,\,::\,\,\,\left\{ \matrix{ \,500 < N\,\,{\mathop{\rm int}} < 2500\,\,\,\,\left( * \right) \hfill \cr \,{N \over {2 \cdot {3^2}}} = {\mathop{\rm int}} \,\,\,;\,\,\,{N \over {3 \cdot {5^2}}} = {\mathop{\rm int}} \,\,\,\,\left( {**} \right) \hfill \cr} \right.$$

$$\left( {**} \right)\,\,\,\, \Rightarrow \,\,\,\,N = k \cdot LCM\left( {2 \cdot {3^2};3 \cdot {5^2}} \right) = k \cdot 2 \cdot {3^2} \cdot {5^2} = 450 \cdot k\,\,\,\,\,\left( {k\,\,{\mathop{\rm int}} } \right)$$

$$\left\{ \matrix{ \,450 \cdot 2 = 900 \hfill \cr \,\left( {450 \cdot 2} \right) \cdot 3 = 2700 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,k = 2,3,4,5$$

$$? = 450\,\left( {2 + 3 + 4 + 5} \right)\,\,\, = \,\,\,6300$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Re: The sum of all integers between 500 and 2500 that are divisible by bot  [#permalink]

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03 Apr 2019, 23:18
1
Another approach will be

Factorise 18 and 75
18 = 3^2 * 2
75 = 5^2 * 3

Therefore LCM = 2*3^2*5^2 ... = 450

The numbers in the series should be between 500 and 2500. So
The first number won't be 450
Let's multiply it by 2 .. we get 900
For the last number, simply divide 2500 by 450.. i.e. 2500/45. The nearest integer will be 5

So total numbers in the series are (5-1)=4
Now the sum formula

Sum = (((First number) + (Last number))*number of terms in the series ) / 2
= (((900 + 2250)*4)/2)
=(1675 * 4)/2
= 6300

So the option B - 6300
Re: The sum of all integers between 500 and 2500 that are divisible by bot   [#permalink] 03 Apr 2019, 23:18
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