Let S be the set of all positive integers having at most 4

[BTGmoderatorDC]

Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

A. 11
B. 19
C. 37
D. 59
E. 101



OA E

Source: Official Guide

[deloitte247]

S = {4-digits positive integer}
Each digit of the element in S is either 0 or 1.
Therefore, set S = {0000, 0001, 0010, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1101, 1110, 1111}
The sum of all elements in set S = 8888
Prime factors of 8888 = 2 * 2 * 2 * 11 * 101
$$=2^3\cdot11\cdot101$$
The greatest prime factor = 101
Answer = option E

[GMATGuruNY]

Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

A. 11
B. 19
C. 37
D. 59
E. 101

Let S = the set of 4-digit integers between 0000 and 1111, inclusive, such that each digit is either 0 or 1.

Number of options for the thousands place = 2. (0 or 1)
Number of options for the hundreds place = 2. (0 or 1)
Number of options for the tens place = 2. (0 or 1)
Number of options for the units place = 2. (0 or 1)
To combine these options, we multiply:
2*2*2*2 = 16 integers

Since each of the 16 integers has 4 digits, there are 16 thousands digits, 16 hundreds digits, 16 tens digits, and 16 units digits.
Each digit has an equal chance of being 0 or 1.
Thus:
Eight of the 16 thousands digits will be 0, while the other eight will be 1, with the result that the sum for the thousands place = 8*0000 + 8*1000 = 8000
Eight of the 16 hundreds digits will be 0, while the other eight will be 1, with the result that the sum for the hundreds place = 8*000 + 8*100 = 800
Eight of the 16 tens digits will be 0, while the other eight will be 1, with the result that the sum for the tens place = 8*00 + 8*10 = 80
Eight of the 16 units digits will be 0, while the other eight will be 1, with the result that the sum for the units place = 8*0 + 8*1 = 8
Resulting sum:
8000 + 800 + 80 + 8 = 8888

8888 = 8 * 1111 = 8 * 11 * 101
Thus, the greatest prime factor for the sum = 101.

The correct answer is E.

[Scott@TargetTestPrep]

Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

A. 11
B. 19
C. 37
D. 59
E. 101

OA E

Source: Official Guide

We see that the numbers in set S are:

1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110 and 1111

We see that there are 15 numbers in set S. Adding the first 7 numbers, we have 1 + 10 + 11 + 100 + 101 + 110 + 111 = 444. Since each of the last 7 numbers is 1000 more than its counterpart in the first 7 numbers, the sum of the last 7 number is 444 + 7 x 1000 = 7444. Now, add 444, 7444 and also the middle number 1000 (which we haven't included in either of the two earlier sums), and we have the sum of the 15 numbers as:

444 + 7444 + 1000 = 8888

Let's now prime factorize 8888:

8888 = 88 x 101 = 8 x 11 x 101 = 2^3 x 11 x 101

We see that the largest prime factor is 101.

Answer: E