The sum of six consecutive multiples of 3 is three times the sum of

Question

The sum of six consecutive multiples of 3 is three times the sum of five consecutive multiples of 5. If the smallest of the six multiples of 3 lies between 90 and 150, inclusive, how many such sets of eleven numbers satisfy this condition?

A. 1
B. 2
C. 3
D. 4
E. 5

A. 1
B. 2
C. 3
D. 4
E. 5

paragw · Answer

Here's my solution

chetan2u · Answer

The sum of six consecutive multiples of 3 is three times the sum of five consecutive multiples of 5 .

Let the first multiple of 3 be x, so numbers are x, x+3, x+6, x+9, x+12 and x+15. Therefore, the sum is 6x+45.

Now, let the first multiple of 5 be y, so numbers are y, y+5, y+10, y+15 and y+20. Therefore, the sum is 5y+50.

Thus,
 6x+45 = 3*(5y+50) ...
 3(2x+15) = 3*5*(y+10) 
 2x+15=5y+50 ....
 2x=5y+35 
or 2x=  5  (y+7)
 So we can say x is multiple of 5. As x is also a multiple of 3, x must be a multiple of 15

Now we know that smallest of the six multiples of 3 lies between 90 and 150, inclusive.

So, we look for multiple of 15   from 90 to 150. 
 The possibilities are 90, 105, 120, 135 and 150.

However, we require to check for which value of x is y a multiple of 5 as we have taken y to be the smallest multiple of 5.

2x=180=5(y+7)....y=29 Not possible
2x=210=5(y+7)....y=35  possible 
2x=240=5(y+7)....y= 41 Not possible
2x=270=5(y+7)....y= 47 Not possible
2x=300=5(y+7)....y= 53 Not possible

Thus, one case.

A

SageCrew · Answer

I think the given solution is wrong. IMO it should be A.

Say the sequence of 3 multiples is 3[k, k+1, k+2, k+3, k+4, k+5] => Sum1: 3[6k+15]

And say the sequence of 5 multiples is 5[l, l+1, l+2, l+3, l+4] => Sum2: 5[5l+10]

Now, given that Sum1 = 3 x Sum2.

Applying:
3[6k+15] = 3 x 5[5l+10]
=> 6k+15 = 25l+50
=> 6k = 25l+35 -------((Eqn 1))

Now, another condition given is that the smallest 3 multiple, 3k, satisfies 90 <= 3k <= 150.
This implies 30 <= k <= 50.
This further implies 180 <= 6k <= 300.
Replacing the above inequality with ((Eqn 1)), we get: 180 <= 25l+35 <= 300

This gives us 145 <= 25l <= 265.
Further simplifying, we get 29 <= 5l <= 53.

So, whatever our set of solutions for both k & l, they must satisfy these inequalities:
30 <= k <= 50 (Therefore the set of possible values for k are 30, 31, 32 ... 50)
29 <= 5l <= 53 (Therefore the set of possible values for l are 6, 7, 8, 9, 10)

Now, we need to find values of k & l such that they satisfy ((Eqn 1))
The only such pair of values are (l = 7 and k = 35)

I think the one mistake that the prior solutions are making is they're treating the (90 <= 3's smallest multiple in the series <= 150) condition as applicable only to the 3-multiple series' values, and they're ignoring that this condition actually also indirectly bounds the 5-multiple series' values, as both the series are connected via the Sum1 = 3 x Sum2 relation.

chetan2u · Answer

Absolutely correct. Kudos

However, the pair you have mentioned is wrong.
The pair as per your solution should be l=7 and k=35.

SageCrew · Answer

Edited, thanks for the correction.

TheAdmissionHub · Answer

We can set up the following equation:
 x+x+3+x+6+x+9+x+12+x+15 = 3* (y+y+5+y+10+y+15+y+20) 
 6x+45=15y+150
6x=15y+105
15y=6x-105
y=\frac{2}{5}x-7

We can see that x has to be multiple of 5 to obtain integer y. Since x is also multiple of 3, x has to be multiple of 15.
Therefore, between 90 and 150 inclusive, x can be: 90, 105, 120, 135, 150.

Since we know that y has to be multiple of 5, let's check the values of x in the equation.

1)  y = \frac{2}{5}*90-7=29  ⭢ not possible

2)  y = \frac{2}{5}*105-7= 35  ⭢  possible

3)  y = \frac{2}{5}*120-7=41  ⭢ not possible

4)  y = \frac{2}{5}*135-7=47  ⭢ not possible

5)  y = \frac{2}{5}*150-7=53  ⭢ not possible

Answer: A

The sum of six consecutive multiples of 3 is three times the sum of

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The sum of six consecutive multiples of 3 is three times the sum of

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