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

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

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New post 21 Sep 2019, 14:02
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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
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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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New post 23 Sep 2019, 06:53
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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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Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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New post Updated on: 22 Sep 2019, 02:56
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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
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Originally posted by Mohammadmo on 21 Sep 2019, 22:12.
Last edited by Mohammadmo on 22 Sep 2019, 02:56, 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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New post 22 Sep 2019, 01:37
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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
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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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New post 23 Sep 2019, 19:20
Mohammadmo 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?
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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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New post 17 Oct 2019, 13:54
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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
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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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New post 20 Oct 2019, 08:27
realslimsiddy wrote:
Mohammadmo 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.

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

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New post Updated on: 13 Nov 2019, 18:46
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


Mohammadmo 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



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!
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Originally posted by dabaobao on 25 Oct 2019, 09:15.
Last edited by dabaobao on 13 Nov 2019, 18:46, edited 2 times in total.
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Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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New post 25 Oct 2019, 10:27
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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Re: Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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New post 29 Oct 2019, 21:54
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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.

Answer (E)
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Let S be the set of all positive integers having at most 4 digits and  [#permalink]

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New post 13 Nov 2019, 18:46
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


Mohammadmo 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



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

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New post 15 Nov 2019, 19:37
VeritasKarishma 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


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.

Answer (E)


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

Adding all of them:
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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Let S be the set of all positive integers having at most 4 digits and   [#permalink] 15 Nov 2019, 19:37
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