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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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{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

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

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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

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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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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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]
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

GIVEN:
• S is a set of ALL positive integers having at most 4 digits.
• Set S can have 1-digit, 2-digit, 3-digit, and 4-digit integers.
• Each digit of an integer in set S is either 0 or 1.

TO FIND:
• The greatest prime factor of the sum of all integers in S.
• Say, the sum of all integers of set S is ‘X’. Then, we need to find the greatest prime factor of X.

APPROACH:
Based on the CONSTRAINT that each digit can be either 0 or 1, first let us find all the integers in S (that is, all 1-digit, 2-digit, 3-digit, and 4-digit integers that can be formed using 0 and 1 as digits). We will then proceed to find X, followed by its prime factors.

WORKING:

Single digit integers in S:
As the only possibilities for the digits of any integer in S are 0 and 1, the only possible single-digit integers in set S are 0 and 1 themselves.
- BUT the question also states that set S has positive integers. Hence, 0 is also not a possibility. Thus, the only single-digit integer in set S is 1.

IMPORTANT: Since our goal is to find the sum of all integers in S, it will be very helpful to have the same number of digits in all the integers. We will rewrite each integer we get using 4 digits, filling 0s when we do not have 4 digits already. (You will see how simple addition becomes once we get to it – Hang on and trust the process! 😊)

So, 1 can be rewritten this way:

2-digits integers in S:
Possible digits for ten’s place are 0 and 1.
Observe: Can a 2-digit number have 0 at its tens place? NO! Because 01 is again just the single-digit number 1!
- So, to have a 2-digit integer, the tens digit can be filled in only ONE way: Ten’s digit = 1.
- Now, for the one’s place, both 0 and 1 are possible. So, one’s place can be filled in TWO ways.

Therefore, total number of 2-digit integers in S = 1 × 2 = 2. These are 10 and 11.

And again, the 4-digit representation of both these numbers are:

3-digits integers in S:
Possible digits for hundred’s place are 0 and 1. But again, just as we saw in the 2-digit case, the hundred’s digit cannot be 0.
Hence, the hundred’s place can be filled in only ONE way: Hundred’s digit = 1.
Then, the remaining two places, ten’s place, and one’s place, EACH can be filled in two ways (by 0 or 1). Therefore, total number of 3-digits integers in S = 1 × 2 × 2 = 4. Check out the possibilities below:

So, the 3-digit integers are 100, 101, 110, 111. And again, we’re going to take these to their 4-digit representations. Here we go:

4-digits integers in S:
Same analysis as above will confirm that the only possible digit for thousand’s place is 1. And the remaining three places, hundred’s place, ten’s place, and one’s place, each can be filled in two ways (by 0 or 1). Therefore, total number of integers = 1 × 2 × 2 × 2 = 8. Check out all possibilities below:

Now, we can easily see that the 4-digit integers are 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111. Since they are already 4-digit integers, we do not need any extra work to write them in their 4-digit representations. Let me just re-write them just as I did for all above integers:

Sum of all integers = X:
In total, set S has a total of 15 integers. That is,

Observe that at every place, the digit ‘0’ comes 7 times (marked in different colors at each place) and ‘1’ comes 8 times.
- Note: If we had included 0 as a single-digit integer, then we would have had 16 total numbers in S. Then, the number of 0s and the number of 1s at every place would have been 8 each. That is, you would have had half 1s and for half 0s at every place.

For our situation, when you try to add the digits at each place, you get: 7 times (0) + 8 times (1) = 8. So, the digit at each place in the sum, X, is 8.
• Hence, X = 8888.

Required prime factored form of X:
The prime factored form of X: 8888 = $$2^3$$ × 11 × 101.
So, the GREATEST PRIME FACTOR of X is 101.

Hope this helps!

Best,
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