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Set A consists of three consecutive positive multiples of 3, and set B

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Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   28 May 2020, 00:03
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Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020
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B
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   28 May 2020, 00:17
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rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020


Let Set A be {\(3x-3, 3x, 3x+3\)}, so sum will be 9x, and Set B be {\(5y-10, 5y-5, 5y, 5y+5, 5y+10\)}, so sum will be 25y.

Given that 9x=25y, so the least value of x will be 25 => \(9*25=25*9\)

SO x=25, and the least number = \(3x-3=3*25-3=72\)

B
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Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   28 May 2020, 10:42
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rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020


Given: Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5.
Asked: If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer

3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
\(k = \frac{(25m+41)}{9}\)
When m = 9; \(k = \frac{(175 + 41 = 216)}{9} = 24\)
\(k_{min} = 24\)
\(3k_{min} = 72\)

IMO B
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   28 May 2020, 12:09
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rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020


The other two approaches mentioned are the perfect way to solve !!

We can also use POE also to get the answer:

Given:
Set A consists of three consecutive positive multiples of 3
set B consists of five consecutive positive multiples of 5
the sum of the integers in set A = sum of the integers in set B


Infer: Set A and Set B are evenly spaced sets
Thus, Mean of the set = Median of the set
Sum = number of items* Mean
Sum of Set A = 3 * Mean of Set A
Sum of Set B = 5 *Mean of Set B
=> 3 * Mean of Set A = 5 *Mean of Set B
=> Mean of Set A is multiple of 5
=> Median of Set A is multiple of 5
Question is asking for :
the least number that could be a member of set A

(A) 69 Not Possible as the median is 72, not a multiple of 5
(B) 72 {72, 75, 78 } Possible, as the median is 75, a multiple of 5. Let's hold it
(C) 75 Not Possible as the median is 78, not a multiple of 5
(D) 78 Not Possible as the median is 81, not a multiple of 5
(E) 81 Not Possible as the median is 84, not a multiple of 5

As we are left with only answer choice B, that's the right answer
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   28 May 2020, 12:34
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The three numbers in set A should have a sum divisible by 5.

A. 69 + 72 + 75 = 216- rejected
B. 72 + 75 + 78= 225 - keep
C. 75 + 78 + 81= 234- rejected
D 78 + 81 + 84 = 243 - rejected
E 81 + 84 + 87 = 252- rejected

Ans 72, B

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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   01 Jun 2020, 08:00
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rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81


We can let a = median of set A and b = median of set B. Then the sum of the 3 integers in set A would be 3a and the sum of the 5 integers in set B would be 5b. Since they are equal, we have:

3a = 5b

We see that a must be a multiple of 5 since 3 is not a multiple of 5. Since a is a multiple of 5, then (a - 3), the smallest member in set A, must be 3 less than a multiple of 5. Looking at the answer choices, we see that only 72 satisfies the condition because it is 3 less than 75 (which is a multiple of 5). Therefore, 72 could be the least number in set A.

Answer: B
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   10 Aug 2021, 22:39
rakshitsood wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020

3*3x=5*5y
since the number chosen in 5 multiples were 5y-10, 5y-5, 5y, 5y+5, 5y+10
and 3 multiples are 3x-3 ,3x,3x+3
=>9x=25y
=>therefore x=25 and y=9
and the least number to satisfy the same is 75-3=72
Threfore IMO B
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   19 Aug 2021, 07:21
rakshitsood wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020


If SUM of B must be equal to SUM of A, then the sum of all the A items MUST BE a multiple of 5.

We thus can go reversal by plugging- in the answers in set A and see whether the sum is a multiple of 5.

IF 72 is the lowest term of A, then set A is (72, 72+3, 72+(2*3)) = 72,75,78

You don't have to calculate the sum, just see if the unit digit of the sum is a 5 or a zero.

Do the same for the other answers and you see that B is the only pattern of which sum gives a number multiple of 5
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   22 Aug 2021, 11:50
Here is a 30 second solution to this problem.

Take a look at the options. For the sum to equal we need the sum of three numbers in Set A to be divisible by 5.

So the units digit can be 5 or 0. Nothing else.

Check the options. A+B+C = Nope
B+C+D = Yuppp
C+D+E = Nope

No other combination will be possible as they are consecutive.
Out of B,C,D the smallest is (B) 72
There's your answer.

Kudos please, if I helped :)
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Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   04 Apr 2024, 09:08
Kinshook wrote:
rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020


Given: Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5.
Asked: If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer

3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
\(k = \frac{(25m+41)}{9}\)
When m = 9; \(k = \frac{(175 + 41 = 216)}{9} = 24\)
\(k_{min} = 24\)
\(3k_{min} = 72\)

IMO B


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Why did you take m=9 and not m=1? Because its the least value that has been asked for. I put m=1 and got k=33. Where am I wrong?
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink] New post   05 Apr 2024, 23:33
Expert Reply
Budhaditya_Saha wrote:
Kinshook wrote:
rsvp wrote:
Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020

Given: Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5.
Asked: If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer

3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
\(k = \frac{(25m+41)}{9}\)
When m = 9; \(k = \frac{(175 + 41 = 216)}{9} = 24\)
\(k_{min} = 24\)
\(3k_{min} = 72\)

IMO B

Posted from my mobile device


Why did you take m=9 and not m=1? Because its the least value that has been asked for. I put m=1 and got k=33. Where am I wrong?

­Actually, m can't be 1 or 9. I believe Kinshook meant to test m=7, since 25(7) = 175. We need a value that produces a multiple of 9 in the numerator, and m=7 is the lowest one that produces that result. 

If you test m=1, you shouldn't get 33. k = (25*1 + 41)/9  = 67/9 = 7r4 (not an integer). 
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Re: Set A consists of three consecutive positive multiples of 3, and set B [#permalink]
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