In list L above, there are 3 positive integers, where each

Karishma - 2 questions
1) How will we find out all the possible values of (A+B+C) in 2mns ?
2) A, B and C are all even. 2+4+8 = 14. Factors 2 and 7
In this case :
111*(A + B + C) should be having the following factors 1, 3, 37 and 111 and 2,7
So sum of factors = 152+2+7=161
So are we not getting 2 ambiguous answers as per the question ?
Please correct me if I am wrong.

Getting 111*(A + B + C) is less than a minute job. After this, we are looking for 2 cases such that (A+B+C) share no common factor. Why? Because the question specifically says 'MUST be factors'. If I can quickly find two cases where (A+B+C) do not have any common factors, I know there is no other factor which MUST be a factor of the sum of the list. Then I will know that only the factors of 111 MUST be factors of the sum of the list. To find the two cases, I will try and take the values of A, B and C as varied as possible in the two cases.
Case 1: All odd to get odd A+B+C = 1+3+5 = 9
Now in the next case I don't want 3 to be a factor. I try to find some even values 2+4+6 = 12 NO. 2+4+8 = 14 Yes. 3 is not a factor here. So there is no other number that definitely needs to be a factor of the sum of the list.



No, in this case, there will be many other factors. 6, 21 etc... Anyway, the question says "MUST be factors". It is not NECESSARY that 2 and 7 will be factors of A+B+C (as shown above in the case where A+B+C = 9). Hence, they will not be included.

The list L consists of :
ABC, BCA, CAB.

Sum of these numbers will be \(111(A+B+C)\).
Since A, B, C are distinct non-zero digits, hence minimum value of A+B+C=6. Hence the sum will be 111*6 or 111*7 or 111*8.........

So atleast \(111\) will be a factor of the sum of the integers.

On prime factorization, we will get 37 and 3 as the prime factors.

Sum of the factors:
{a^ (p+1) - 1}{b^ (q+1) - 1}{c^ (r+1) - 1} / (a-1)*(b-1)*(c-1)

Here a=37, b=3, p=1, q=1

On applying the formula:
we get \((1368*8)/(36*2)\)
or \(152\).
Hope that helps.

To explain still elaborately
take any number of the format given:123+231+312=666==>111(1+2+3)
So now the question asks for which of the following will be the factors of all the three postive integers.
Only 111 can be the factor of all 3 +ve integers
So the factors of 111 are 1,3,37,111..
Making a sum ofall the factors gives the answer E

KAPLAN OE :

The question is very involved and somewhat convoluted sounding! It asks us about distinct digits that are arranged in different orders to form 3 different numbers, which tells us that the digits each variable represents will remain constant, but each number’s value will change depending on the position each digit occupies (units, tens, hundreds). We need to determine the sum of all the factors of the sum of the three numbers. Since the question asks about those integers that must be factors, we should be wary of finding integers that could be factors. This question is testing both our calculation skills and knowledge of the position of digits in numbers.

Identify the Task:

We need to determine the integers in the list, their sum, and the factors of that sum.

Approach Strategically:

The calculations involved will make Backsolving a challenge. Since the question asks which integers must be factors, if we decide to Pick Numbers, we should be aware that we could wind up finding integers that are factors only with the values we pick, and as such we might need to pick multiple sets of numbers. This strategy is certainly a possibility, but doing the Straightforward Math will be our fastest option.

The first digit to the left of the decimal represents the ones or units’ digit. The digit to the immediate left of the units’ digit is the tens’ digit (and its value is 10 times whatever digit is in that place), and the digit to the immediate left of the tens’ digit is the hundreds’ digit (and its value is 100 times whatever digit is in that place). Therefore the value of the integer ABC is equal to 100A + 10B + C; the value of the integer BCA is equal to 100B + 10C + A; and the value of the integer CAB is equal to 100C + 10A + B.

The sum of the integers ABC, BCA, and CAB can therefore be expressed as:

, which equals 111(A + B + C)

The sum of the integers in list L must be a multiple of 111, but depending on the value of (A + B + C) it could be a multiple of several other numbers. Any positive integer that is a factor of 111 also must be a factor of the sum of the integers in list L. 111 = 3 × 37, and both 3 and 37 are prime numbers. So the positive integers that are factors of 111 are 1, 3, 37, and 111. When we sum these, we get 1 + 3 + 37 + 111 = 152. Answer Choice (C) is correct.

Confirm your Answer:

We can also Pick Numbers to verify this. The sum of the numbers in list L is 111(A + B + C). If A = 1, B = 2, and C = 4, then A + B + C =1 + 2 + 4 = 7. With these values, the sum 111(A + B + C) of the numbers in list L would have the factors: 1, 3, 37, 111, and 7.

However, If A = 1, B = 2, and C = 5, then A + B + C =1 + 2 + 5 = 8. With these values, the sum 111(A + B + C) of the numbers in list L has the factors 1, 3, 37, 111, and 8. Neither 7 nor 8 MUST be a factor of the sum, but 1, 3, 37, and 111 always must be.

If we picked numbers from the beginning, we would see something similar. If A = 1, B = 2 and C = 4, our list would be comprised of 124, 241 and 412. The sum would be 777. The prime factors would therefore be 3, 7 and 37, and all of the factors would be 1, 3, 7, 21, 37, 111, 259, 777. Without picking another set of numbers we could not easily determine which of these must be factors of the sum.

If A = 1, B = 2 and C = 5, our list would be comprised of 125, 251, and 512. The sum would be 888. The prime factors would be 2, 2, 2, 3 and 37 and all of the factors would be 1, 2, 3, 4, 6, 8, 12, 24, 37, 74, 111, 148, 222, 296, 444, 888. Only 1, 3, 37 and 111 are common to both lists.

The 3 Digit Integers expressed in Power of 10 “Notation” Form, where A, B, and C are Digits

ABC = 100A + 10B + 1C

BCA = 100B + 10C + 1A

CAB = 100C + 10A + 1B
_________________________

Adding the 3 Numbers =

111A + 111B + 111C =

111 * (A + B + C)

Regardless of the Digits that fill the A , B , and C Spots, the SUM of the 3 Numbers Must Always be Divisible by 111 and all the Factors of 111


Unique +Pos. Factors of 111
____________
1 ; 3 ; 37 ; 111

1 + 3 + 37 + 111 =

152

-C-

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