If the sum of 5 consecutive positive multiples of a prime number is 10

So from the question, let p be the prime number.

p*x + p*(x+1) + p*(x+2) + p*(x+3) + p*(x+4) = 105
5p*x + 10p = 105
5p*(x + 2) = 105
p*(x+2) = 21

Now 21 can be written as 3*7 or 7*3

Case 1: 21 = 3*7 then p = 3, x + 2 = 7 or x = 5
Then, sum of 1st 5 positive multiples of 3 = 3 + 6 + 9 + 12 + 15 = 45

Case 2: 21 = 7*3 then p = 7, x + 2 = 3 or x = 1
Then, sum of 1st 5 poistive multiples of 7 = 7 + 14 + 21 + 28 + 35 = 105

Since we get 2 sums, the answer can't be determined.

Option E!

Answer D
sum of 5 multiples of 7- 7 14,21, 28, 35 , they are the 1st 5 multiples of 7 as well, sum is 105

E.
Consecutive 5 multiples means that the product will be of form (1+2+3+4+5)*Prime Number =105.

Now since 105 is odd...so 5 consecutive numbers have to start in odd like 12345 or 34567 etc.

Also tha factors of 105 are 3 5 and 7.
Taking them as prime number we get summation as

Prime mumber = 7
Summation =15

Prime number = 3
Summation = 35


So both cases are possible hence cant be determined.

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Let the prime number be x
5 consecutive multiple of x- x,2x,3x,4x,and 5x
Their sum is given as 105
x+2x+3x+4x+5x=105
15x=105
x=7
Sum of first 5 positive multiple of 7=7+14+21+28+35
=105

Posted from my mobile device

Since the sum is 105

Prime number factors: 3,5,7

Let the prime number be 3

Thus the consequetive numbers are

(2x +1) +(2x +4) +(2x +7) +(2x +10) +(2x +13)= 105
10x+35=105
x= 7
Numbers: 15, 18, 21, 24 & 27

Let the prime number be 7

Thus the consequetive numbers are

(2x+5) +(2x +12) +(2x +19) +(2x +26) +(2x +33)= 105
10x+95=105
x= 1
Numbers: 7, 14, 21, 28 & 35

Hence 2 such possibilities.

Cant be determined

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Sum of 5 consecutive positive multiples of a prime number is 105 .............. Given

Using this property, lets assume this prime number as N and first multiple out of 5 consecutive ones as X.

Adding 4 consecutive multiple of N to X for the sum, we get -
X + X + N + X + 2N + X + 3N + X + 4N = 105
that is, 5X + 10N = 105
that is, X + 2N = 21 ................. (I)

Now, N cannot be 2 as sum of it's 5 multiples is odd and sum of any multiple of 2 is even and N cannot be greater than 10 otherwise 2N will be greater than 21 making above equation invalid.

So, range of N is between 3 and 10, inclusive and possible values of N are 3,5 and 7.

We know that X is a multiple of N. Using (I) from above, start plugging in values in reverse.

N = 7 gives 2N = 14 and X = 6 - Incorrect as 6 is not multiple of 7.
N = 5 gives 2N = 10 and X = 11 - Incorrect as 11 is not multiple of 5.
N = 3 gives 2N = 6 and X = 15 - CORRECT as 15 is multiple of 3.

So, N = 3 and sum of first 5 multiples of N = 3+6+9+12+15 = 45.

Hence, B is the correct answer.

C is the answer.

Sum is 105, two suspects are 5 and 7. If we try with 7, even with first five integers, the sum amounts to 108. What left is 5, starting from 15 and ending at 35, sum amounts to exactly 105. Adding first 5 consecutive multiple of 5 gives you 75.

Thanks,

If five consecutive multiples of the number sum to 105, then since we have an odd number of integers, the mean and median of the set of numbers must be 105/5 = 21.

Looking at 21, we can see that it has two prime factors of 7 and 3, therefore 3 and 7 could both have sets of consecutive multiples with a middle term of 21 (7, 14, 21, 28, 35 OR 15, 18, 21, 24, 27.) Because we can't determine if the prime number is 3 or 7, we can't determine the value of the first five multiples, since it could be the first five multiples of 3 or the first five multiples of 7.


E

Let Prime No. =K & 5 Consequtive multiples be x, x+1, x+2, x+3, x+4

A/Q, Kx + K(x+1) + K(x+2) + K(x+3) + K(x+4)=105
=> 5Kx + 10K = 105
=> K(x+2)= 21= 7*3
So either K=3 {x=5} or K=7 {x=1}
So two possibilities for the sum of products.

Ans E

Two ways

I.
Let the prime number be p, then sum of any 5 consecutive multiples can be p*t, p*(t+1) and so on.
Sum of the first five multiples of p=p+2p+3p+4p+5p=15p

Sum of these multiples = \(p*t+p*(t+1)+p*(t+2)+p*(t+3)+p*(t+4)=105\)
\(5tp+p(1+2+3+4)=105……….5tp+10p=105………(t+2)p=21=3*7\)

Cases
a) p=3, t+2=7 or t=5
So 3*4,3*5…..,
15,18,21,24,27
SUM=15p=15*3=45

b) p=7, t+2=3 or t=1
So 7*1,7*2……
7,14,21,28,35
SUM=15p=15*7=105

As p can have two different values, we can have different SUMs.


II. Median

Since 105 is sum of 5 consecutive multiples, the median will be 105/5=21=3*7
Cases
a) 3*7 is median
3*5, 3*6, 3*7, 3*8, 3*9

b) 7*3 is median
7*1, 7*2, 7*3, 7*4, 7*5

Different answers possible



E
_________________

Didn't do this algebraically. That looks too complicated for me.

Lets take the prime numbers 2,3,5,7,11 etc

First checked Multiples of 5 - 5,10,15,20,25,30,35 - couldn't get 105 with 5 consecutive multiples.

Then checked 7 - 7,14,21,28,35 (here 21*5 = 105 - Even spaced so mean*no of integers).

Now can there be any other 5 consecutive multiples with 21 as mean. Yes, the multiples of 3 i.e. 15,18,21,24,27.

The sum of first 5 multiples of 7 is 105 but the sum first 5 multiples of 3 is obviously not 105.

So, answer is E - can't be determined.