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How I did it- Please let me know if I my thought process is incorrect

Let the prime number be p
5 consecutive multiples for p would be pa, p(a+1), p(a+2),p(a+3), p(a+4)

Sum of 5 consecutive multiples of p=
=p[a+(a+1)+(a+2)+(a+3)+(a+4)]
=p[5/2(2a+4)] (Sum of AP)

105=p[5/2(2a+4)]

Thus we will not be able to determine p and a using only one equation .
Hence cannot be determined
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Hi,
Choose a prime number - 5
5 + 10+ 15 + 20 + 25 = 75
Ans could be c

15 + 20 + 25 + 30 + 35 > 105
10 + 15 + 20 + 25 + 30 > 105

Choose another prime number - 7
7 + 14+ 21 + 28 + 35 = 105 Ans D
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niteshwaghray
If the sum of 5 consecutive positive multiples of a prime number is 105, what is the sum of the first 5 positive multiples of that prime number?


A. 30
B. 45
C. 75
D. 105
E. Can't be determined


Hi,

Let the no. be P.
As per the question, \(P*x + P*(x+1) + P*(x+2) + P*(x+3) + P*(x+4) = 105\) --- eqn. 1
Solving the equation, we get :

\(5*P*x + 10*P = 105\\
=> P*x + 2*P = 21\\
=> P(x+2) = 21\)

Factors of 21 - 3,7,21

So\(P*(x+2)=21\) can either be \(7*(1+2)\) or\(3*(1+6).\)

Lets try the values : If x=1 and P=7, substituting in eqn. 1, we get \(7+14+21+28+35 = 105\)--- Satisfied
but if we consider P=3 and x=6, substituting in equation leads to terms - \(18+21+24+27+30 = 120\) which doesn't suffice.

Thus, P must be equal to 7 and x must be equal to 1.

Based on these, the sum of 1st 5 multiples = \(7+14+21+28+35 =105\)


As per me, option D should be the answer, can somebody please suggest why is it E ?

Got it. :oops: Such a careless mistake.
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Imo E
As we have two unknowns hence we can not determine sum of multiples .
Let the prime number be X and and Y be some positive integer .
Now according to question we have , X[y+(y+1)+(y+2)+(y+3)+(y+4)
= X[5y+10]
Thus we have two unknowns that can not determine answer



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niteshwaghray
If the sum of 5 consecutive positive multiples of a prime number is 105, what is the sum of the first 5 positive multiples of that prime number?


A. 30
B. 45
C. 75
D. 105
E. Can't be determined


Let the prime number be x, and so sum of five positive multiples will be
x*(y-2)+x(y-1)+xy+x(y+1)+x(y+2)=105
x(y-2+y-1+y+y+1+y+2)=105
x*5y=105......x*y=21
Since x is prime, 21 can be factored as only 7*3. Thus x can be 7 or 3.
Insufficient information

R
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two sets are possible for the question.
3: 15 18 21 24 27
7: 7 14 21 28 35

Hence the answer is can not be determined for the provided info.

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niteshwaghray
If the sum of 5 consecutive positive multiples of a prime number is 105, what is the sum of the first 5 positive multiples of that prime number?


A. 30
B. 45
C. 75
D. 105
E. Can't be determined

Consecutive multiples constitute an EVENLY SPACED SET.
For any evenly spaced set, median = average.
Given 5 consecutive multiples that sum to 105, we get:
median = average =\( \frac{sum}{count} = \frac{105}{5} =21\)

21 is a multiple of prime numbers 3 and 7.

Case 1: 5 consecutive multiples of 3 such that the median is 21
15, 18, 21, 24, 27 --> 15+18+21+24+27 = 105

Case 2: 5 consecutive multiples of 7 such that the median is 21
7, 14, 21, 28, 35 --> 7+14+21+28+35 = 105

The sum of the first five multiples of 3 is not equal to the sum of the first five multiples of 7.
Thus, the answer to the question stem cannot be determined.

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