In a finite list of positive integers, each number, except for the las

Question

In a finite list of positive integers, each number, except for the last number, is a proper divisor of the next number in the list. If the first number in the list is 1 and the last is 650, what is the greatest possible sum of the numbers in the list?

(A) 753
(B) 892
(C) 934
(D) 1054
(E) 1128

(A) 753
(B) 892
(C) 934
(D) 1054
(E) 1128

GraemeGmatPanda · Answer

Nice question  Ravixxx

Prime breakdown of 650 is  2, 5^2, 13

Starting from the end of the list (650) and working backwards, the highest proper divisor of 650 is 5^2x13 = 325
The next one will be 5x13 = 65
The next one will be 13, which is prime hence we get to the end of the list (with 1 at the end)

Adding them all up, we get to (D) 1,054

TestPrepUnlimited · Answer

The list starts from 1 and ends at 650, for each term as we progress we will add one more factor and eventually hit 650. 650 = 65*10 = 13*5*5*2.
There are many ways to achieve this, we can multiply by 2, then 5, then 5, then 13 and that works our way towards 650. We could arrange the factors in any order but the question wants us to find the arrangement that produces the greatest sum.

To do that, we need to start by feeding the big factors first which will let us approach 650 faster. Then the list would be {1, 13, 65, 325, 650}. This adds up to 1054, which is answer D.

If in doubt that this is the best arrangement, we can always try the other extreme option, which is to multiply smaller factors first. This gives us {1, 2, 10, 50, 650} and we can visually see the numbers are much smaller.

Ans: D

NetOrb · Answer

BUT IT SAYS LAST NUMBER SHOULD NOT BE PROPER DIVISOR, HOW ARE WE INCULCATING THAT?

Bunuel · Answer

In a finite list of positive integers, each number, except for the last number, is a proper divisor of the next number in the list. If the first number in the list is 1 and the last is 650, what is the greatest possible sum of the numbers in the list?

(A) 753
(B) 892
(C) 934
(D) 1054
(E) 1128

(A) 753
(B) 892
(C) 934
(D) 1054
(E) 1128

A proper divisor of a number is a positive divisor that is  smaller  than the number itself. For example, 325 is a proper divisor of 650, but 650 is not a proper divisor of 650.

Every number in the list must divide the next one, so every number in the list must eventually be a divisor of 650.

650 = 2 * 5^2 * 13

To maximize the sum, work backward and choose the largest possible proper divisor each time:

Largest proper divisor of 650 is 325.
Largest proper divisor of 325 is 65.
Largest proper divisor of 65 is 13.
Largest proper divisor of 13 is 1.

So the list can be:

1, 13, 65, 325, 650

Sum:

1 + 13 + 65 + 325 + 650 = 1054

Answer: D.

It does not mean the last number “should not be a proper divisor.” It means the last number is exempt from the rule because there is no next number after it.

KarishmaB · Answer

First number = 1, Last number = 650
 650 = 2*5^2 * 13

Every number should be a proper divisor of the next number. This means that once we bring in a factor, it must be carried forward. 
To get the "maximum" sum, we should start with the greatest prime factor and then keep adding smaller and smaller factors. This will give us larger factors and hence maximum sum.

So the list becomes 
1, 13, 13*5, 13*5*5, 13*5*5*2

Sum = 1 + 13(5 + 25 + 50) = 1054

Answer (D)

Note that if we start with the smallest factor and carry those forward, fewer numbers will have 13, the largest factor. E.g. 1, 2, 2*5, 2*5*5, 2*5*5*13

In a finite list of positive integers, each number, except for the las

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In a finite list of positive integers, each number, except for the las

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