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Senior Manager  P
Joined: 04 Sep 2017
Posts: 318
Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 53% (02:19) correct 47% (02:46) wrong based on 413 sessions

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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

PS41661.01
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10623
Location: Pune, India
Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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5
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gmatt1476 wrote:
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

PS41661.01

We need positive integers having at most 4 digits (so 1 digit, 2 digit and 3 digits are also allowed) such that each digit is 0 or 1. Note that 1 is allowed which is the same as 0001. So basically S is a set of all 4 digit numbers such that any digit can be 0 or 1.

S = __ __ __ __

We can make 2*2*2*2 = 16 such positive integers since we can fill in each of the 4 spaces in 2 ways. Now imagine writing these 16 numbers one below the other to add.

0 0 0 0
0 0 0 1
0 0 1 0
...
1 1 1 1
--------

When we add them, noticing the symmetry we know that there will be 8 0's in units digits and 8 1's. So units digits will add up to 8. Similarly, tens digits, hundreds digits and thousands digits will all add up to 8.
Sum of all numbers in S = 8888

8888 = 8 * 1111 = 8 * 11 *101

101 is the largest prime factor of sum of all numbers in S.

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Karishma
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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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4
1
3
suppose a 4 digit number- 'abcd'
As there is 0 or 1 at each digit place, total possible numbers= 2*2*2*2=16

There are 8 cases when a, b, c or d is 0, and there are 8 cases when those digits are 1

Sum of all possible integers= 8*(1111)= 8*11*101

gmatt1476 wrote:
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

PS41661.01
##### General Discussion
Senior Manager  G
Joined: 28 Jun 2019
Posts: 469
Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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4
1
{0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111}
8*1000+8*100+8*10+8*1=8888
is divisible by 101.
Option E

Posted from my mobile device
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Always waiting

Originally posted by Mohammadmo on 21 Sep 2019, 21:12.
Last edited by Mohammadmo on 22 Sep 2019, 01:56, edited 1 time in total.
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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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1
2
Possible values with digits 0 or 1
1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,111
sum of these digits ; 8*111 ; 2^3*11*101
greatest prime factor ; 101
IMO E

gmatt1476 wrote:
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

PS41661.01
Senior Manager  P
Joined: 01 Feb 2017
Posts: 270
Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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1
4
From 0000 to 1111, there are 4 places and each place can have digits 0 or 1.

So, there will be 16 integers in all and each digit will be repeated 8 times per place.

Hence, sum of digits per place is 8 and total sum of all the integers is 8888.
(even if we are considering positive integers only, ignoring 0000 will not make any difference to this sum of 8888)

Factorizing 8888: 11*2^3*101

Highest PF: 101

Ans E
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Joined: 18 Aug 2019
Posts: 4
Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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{0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111}
8*1000+8*100+8*10+8*1=8888
is divisible by 101.
Option E

Posted from my mobile device

Can you please explain why 8 has been multiplied with 1000, 100, 10 and 1? Also, does that step even depend on the list of possible values you've listed above?
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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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realslimsiddy wrote:
{0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111}
8*1000+8*100+8*10+8*1=8888
is divisible by 101.
Option E

Posted from my mobile device

Can you please explain why 8 has been multiplied with 1000, 100, 10 and 1? Also, does that step even depend on the list of possible values you've listed above?

Hey realslimsiddy,
Since there are 8 1's in the Thousand place, 8 1's in the hundreds place, 8 1's in the tenths place and 8 1's in the Unit's place, 8 has been multiplies by each to represent the number 8888 = 8*1000+8*100+8*10+8
Hope this helps.

Please give Kudos if it helped Director  V
Joined: 24 Oct 2016
Posts: 704
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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gmatt1476 wrote:
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

PS41661.01

{0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111}
8*1000+8*100+8*10+8*1=8888
is divisible by 101.
Option E

Posted from my mobile device

Sum with repetition being allowed is: n^(n−1)∗(sum of the digits)∗(111... n times) => I believe this formula is not valid when one of the digits is 0 since 0 at the 1st spot decreases the # of digits by 1.

Bunuel VeritasKarishma Could you please explain a faster way to do this Q? The above solution would take too long. I can't seem to figure out a shortcut to solve this Q with logic instead of tedious arithmetic. Thanks!

Originally posted by dabaobao on 25 Oct 2019, 08:15.
Last edited by dabaobao on 13 Nov 2019, 17:46, edited 2 times in total.
Director  P
Joined: 04 Aug 2010
Posts: 653
Schools: Dartmouth College
Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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1
gmatt1476 wrote:
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 each integer in S be represented by ABCD, where each digit is either 0 or 1.

Total number of integers:
Number of options for A = 2. (0 or 1)
Number of options for B = 2. (0 or 1)
Number of options for C = 2. (0 or 1)
Number of options for D = 2. (0 or 1)
To combine these options, we multiply:
2*2*2*2 = 16

Digits:
Total number of digits = (16 integers)(4 digits per integer) = 64
Since each of the 64 digits has an equal chance of being 0 or 1, we get:

Eight 1's and eight 0's in the thousands place, implying the following sum for A:
(8*1 + 8*0) * 1000 = 8000
Eight 1's and eight 0's in the hundreds place, implying the following sum for B:
(8*1 + 8*0) * 100 = 800
Eight 1's and eight 0's in the tens place, implying the following sum for C:
(8*1 + 8*0) * 10 = 80
Eight 1's and eight 0's in the units place, implying the following sum for D:
(8*1 + 8*0) * 1 = 8

Resulting sum = 8000 + 800 + 80 + 8 = 8888 = 8*1111 = 8*11*101
The greatest prime factor is the value in blue.

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Director  V
Joined: 24 Oct 2016
Posts: 704
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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dabaobao wrote:
gmatt1476 wrote:
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

PS41661.01

{0,1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111}
8*1000+8*100+8*10+8*1=8888
is divisible by 101.
Option E

Posted from my mobile device

Sum with repetition being allowed is: n^(n−1)∗(sum of the digits)∗(111... n times) => I believe this formula is not valid when one of the digits is 0 since 0 at the 1st spot decreases the # of digits by 1.

Bunuel VeritasKarishma Could you please explain a faster way to do this Q? The above solution would take too long. I can't seem to figure out a shortcut to solve this Q with logic instead of tedious arithmetic. Thanks!

Bunuel Sorry to mention you again. I've seen in many posts that you use the formula mentioned above in red. It doesn't seem to work for this Q. Can you suggest the reason for it? When is it safe to use that formula? For example, we can use the above formula in this Q: https://gmatclub.com/forum/there-are-27 ... ml?kudos=1
Director  V
Joined: 24 Oct 2016
Posts: 704
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Let S be the set of all positive integers having at most 4 digits and  [#permalink]

### Show Tags

gmatt1476 wrote:
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

PS41661.01

We need positive integers having at most 4 digits (so 1 digit, 2 digit and 3 digits are also allowed) such that each digit is 0 or 1. Note that 1 is allowed which is the same as 0001. So basically S is a set of all 4 digit numbers such that any digit can be 0 or 1.

S = __ __ __ __

We can make 2*2*2*2 = 16 such positive integers since we can fill in each of the 4 spaces in 2 ways. Now imagine writing these 16 numbers one below the other to add.

0 0 0 0
0 0 0 1
0 0 1 0
...
1 1 1 1
--------

When we add them, noticing the symmetry we know that there will be 8 0's in units digits and 8 1's. So units digits will add up to 8. Similarly, tens digits, hundreds digits and thousands digits will all add up to 8.
Sum of all numbers in S = 8888

8888 = 8 * 1111 = 8 * 11 *101

101 is the largest prime factor of sum of all numbers in S.

To elaborate a bit further on the above solution:

Method 1: Sum Each Column of Digits (Continuation of above)

So units digits will add up to 8*0 + 8*1 = 8

Similarly, tens digits will add up 8*10
Similarly, hundreds digits will add up 8*100
Similarly, thousands digits will add up 8*1000

8 + 8*10 + 8*100 * 8*1000 = 8 * (1 + 10 + 100 * 1000) = 8 * 1111 = 8888

Method 2: Direct Formula

n^(n−1)*(sum of the digits)*(111... n times) won't work since that's used for Sum of all n digit numbers formed by n non-zero digits
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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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2
1
gmatt1476 wrote:
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

PS41661.01

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.

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Intern  B
Joined: 02 Jun 2020
Posts: 2
Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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gmatt1476 wrote:
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

PS41661.01

We need positive integers having at most 4 digits (so 1 digit, 2 digit and 3 digits are also allowed) such that each digit is 0 or 1. Note that 1 is allowed which is the same as 0001. So basically S is a set of all 4 digit numbers such that any digit can be 0 or 1.

S = __ __ __ __

We can make 2*2*2*2 = 16 such positive integers since we can fill in each of the 4 spaces in 2 ways. Now imagine writing these 16 numbers one below the other to add.

0 0 0 0
0 0 0 1
0 0 1 0
...
1 1 1 1
--------

When we add them, noticing the symmetry we know that there will be 8 0's in units digits and 8 1's. So units digits will add up to 8. Similarly, tens digits, hundreds digits and thousands digits will all add up to 8.
Sum of all numbers in S = 8888

8888 = 8 * 1111 = 8 * 11 *101

101 is the largest prime factor of sum of all numbers in S.

I don't understand why are we considering any number < 1000 as four-digit number. If it would have been a password, the logic is okay, but this question talks about simple numbers
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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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gmatt1476 wrote:
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

PS41661.01

If digits can be 1 or 0 and at most there can be 4 digits, the greatest number that can be formed is 1111
Simply divide 1111 with all options
(A) 11x101
(B) ,(C),(D) not factors
(E) Correct Answer (since we want the greatest one)
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************************************* Re: Let S be the set of all positive integers having at most 4 digits and   [#permalink] 05 Jun 2020, 07:40

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