The three-digit positive integer n can be written as ABC, in which

Question

Tough and Tricky questions: Remainders.

The three-digit positive integer  n  can be written as  ABC , in which  A ,  B , and  C  stand for the unknown digits of  n . What is the remainder when  n  is divided by 37?

(1)  A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}

(2)  A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}

Kudos for a correct  solution .

Bunuel · Accepted Answer

Official Solution:

The question stem tells us that the positive integer  n  has three unknown digits:  A ,  B , and  C , in that order. In other words,  n  can be written as  ABC . Note that in this context,  ABC  does not represent the product of the variables  A ,  B , and  C , but rather a three-digit integer with unknown digit values. It is important to note that since  A ,  B , and  C  stand for digits, their values are restricted to the ten digits 0 through 9. Moreover,  A  cannot equal 0, since we know that  n  is a "three-digit" integer and therefore must be at least 100.
 
We are asked for the remainder after  n  is divided by 37. We could rephrase this question in a variety of ways, but none of them are particularly better than simply leaving the question as is.
 
Statement (1): SUFFICIENT. We can translate this statement to a decimal representation, which will be easier to understand. The left side of the equation, in words, is " A  units plus  B  tenths plus  C  hundredths." We can write this in shorthand:  A.BC  (that is, "A point BC"). After performing the same translation to the right side of the equation, we can see that we get the following:
  A.BC = B.CA 
 
Since  A ,  B , and  C  stand for digits, we can match up the decimal representations and observe that  A = B  and  B = C . Thus, all the digits are the same.
 
This means that we can write  n  as  AAA , which is simply  111 	imes A .
 
Now, 111 factors into  3 	imes 37 , so  n = 3 	imes 37 	imes A . Thus,  n  is a multiple of 37, and the remainder after division by 37 is zero.
 
Statement (2): SUFFICIENT. Again, we can match up the decimal representations of the given equation and find that all the digits are the same. The logic from that point forward is identical to that shown above.

Answer: D

revul · Answer

I found this quite challenging to conceptualize so perhaps this is off:

The three-digit positive integer n can be written as ABC, in which A, B, and C stand for the unknown digits of n. What is the remainder when n is divided by 37?

(1) A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}

(2) A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}

1). This can be rewritten as 100 A + 10 B + C = 100 B + 10 C + A.

Simplify for a value: 99A = 90 B + 9C. I almost gave up there
11A = 10B + C. No value can be negative. All are integers... can A=B=C? There 111, 222, 333, etc. With 111 being 37 * 3 (90+21) then 111, 222, 333 must all be multiples of 37, so remainder is 0

Can we have anything else? Lets make A 2. 22 = 10+2. So we can have 212 as a number. Different remainder. A is insufficient.

2)
rewrite to 100A + 10V + C = 100C + 10A + B --> 90A = 9B +99C... 10A = B+11C.

382 works, 473, 564. These give different remainders, insufficient.

Together...
hold on.

11a = 10b + c and 10a = b + 11c. Can't be negative, both have been disproven, and these say conflicting things. Must be E

harshaiit · Answer

Ground rules: 0<= A,B,C <=9 - to form a 3 digit number.

1) Re-arranging: 11A = 10B + C - Applying the ground rule; A = (10B+C)/11 => 10B+C has to be multiple of 11 starting from 0 to 99. 0 isn't going to give a 3 digit number (A=0 makes it 2 digit). So, 10B+C=11,22....99 and A = 1,2,....9. So the numbers that fit this condition are 111,222,333......999. All these numbers are divisible by 37. So the remainder = 0. A gives an unambiguous answer

2) Re-arranging: 11C = 10A + B - Same as above; C=1,2,....9 and 10A+B = 11,22...99 and the numbers that fit this condition are again 111,222,333.....999. Same as above and B too gives an unambiguous answer.

Answer should be (D). OA?

gillesalex07 · Answer

Hi there,

(1) implies ABC=BCA therefore A=B=C
A is at least 1 so ABC is 111, 222, etc.
111/37 = 3 so the remainder is zero.
Same will be true with 222, etc. as they are multiples of 111.

(2) same reasoning as above

Solution: D

kinghyts · Answer

n = ABC = 100A +10B + C

From statement 1 : A + B/10 + C/100 = B + C/10 + A /100
or, 100A + 10B+C = 100B + 10C + A

From this we can say A=B=C , therefore the possible numbers are 111, 222, 333, 444... , which are all divisible by 37.

From statement 2, we can deduce the same.

Therefore the answer has to be D) , since each of the statements is sufficient.

rdoshi · Answer

From either 1 or 2, we can deduce that

A=B=C

111 = 37*3 so all multiples of 111 (222,333, 444,...,999) will be divisible by 37.

Thus, each statement alone is sufficient. Answer should be D.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The three-digit positive integer n can be written as ABC , in which A , B , and C stand for the unknown digits of n . What is the remainder when n is divided by 37?

When you modify the original condition,

(1) A+B10 +C100 =B+C10 +A100 Multiply 100 to the both equations. It becomes 100A+10B+C=100B+10C+A -> 99A=90B+9C, 11A=10B+C.

(2) A+B10 +C100 =C+A10 +B100 Multiply 100 to the both equations. It becomes 100A+10B+C=100C+10A+B -> 99C=90A+9B, 11C=10A+B.

In the original condition, there are 3 variables, which should match with the number of equations. So, you need 3 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer as 1 more equation is needed. When 1) & 2), in 1)=2), n=111,222,333,.......,999. In 111=37*3+0, 0 is the remainder when n is divided by 37, which is unique and sufficient. Therefore, the answer is D. When 1)=2), it is about 95% that D is the answer.

Ekland · Answer

To solve this math in a CAT u need to stick a couple of math rules to your head otherwise u cant in 3 minutes.
FIRSTLY  234 = 100(2) + 10(3) + 4  
 200 + 20 + 4 = 234 
This works for any number such that the four digit number  ABDE = 1000(A) + 100(B) + 10(D) + E

SECONDLY, the 37 rule: 37 is can divide 111, 222, 333, 444, 555, ..., 999.

Now SOLUTION to the question above
(1) if rearranged, simply states that  ABC = BCA  (i.e. for the 3-digit integer, when u place the first digit as the last and the second as the first it must equal the original number in value)
only if the three digits are equal will that condition hold.
So ABC must equal 111 or 222 or 333 or 444 or 555 or 666 or 777 or 888 or 999.
We dont care which of them it is. Each leaves a remainder 0 when divided by 37.
Sufficient!

For 2, same condition in 1. sufficient.
D. Each.

IanStewart · Answer

If ABC is a three-digit number, where A, B and C are digits, then it is equal to 100A + 10B + C.

Multiplying by 100 on both sides of Statement 1, we find

100A + 10B + C = 100B + 10C + A

But if A, B and C are between 0 and 9, the left side of this equation is just the three digit number ABC, and the right side is BCA. And if ABC and BCA are the same number, they must have the same hundreds digit, so A = B, and the same tens digit, so B = C. So A = B = C, and our number is actually AAA. Since 111 is divisible by 37 (it is 37*3), any three-digit number AAA will be divisible by 37 (it will be 37*(3A)). So using Statement, the required remainder is zero and Statement 1 is sufficient.

Statement 2 tells us

100A + 10B + C = 100C + 10A + B

which means ABC = CAB, and again, A = C and A = B, so our number looks like AAA, and is divisible by 37. So the answer is D.

shridhar786 · Answer

n = 100A+10B+C

STATEMENT 1 --- A+ \frac{B}{10} + \frac{C}{100} =B+ \frac{C}{10} + \frac{A}{100}

100A+10B+C = 100B+10C+A
99A-90B-9C=0
11A-10B-C=0
11A=10B+C 
now n = 100A+11A (putting value of 10B+C)
n = 111A divisible by 37
so sufficient

STATEMENT 2 -- A+ \frac{B}{10} + \frac{C}{100} =C+ \frac{A}{10} + \frac{B}{100} 
100A+10B+C=100C+10A+B
99C-90A-9B=0
11C-10A-B=0
B=11C-10A
now n = 100A+10(11C-10A)+C (putting value of B)
N = 100A+110C-100A+C
N = 111C divisible by 37
so sufficient

ans D

wtstone · Answer

What is stopping me from setting A, B, and C equal to each other... 0,1,2,... infinity...? in that case the answer would be E. I know how I am likely falling in to a trap with that logic or I am understanding/accounting for restrictions, conditions, etc...

All help is greatly appreciated!

All help is greatly appreciated!

G-Man · Answer

Remainder when divided by 37

A). A + B/10 + C/100 = B + C/10 + A/100
This is possible only if all 3 digits are the same (If they are different, the equation will not be correct)
Implies, n = 111, 222, ..... 999
If you check 111, it is a multiple of 37. All following numbers will also be multiples of 37 in that case
A is sufficient

B). It is the same a s A
Hence B is also sufficient

Both are sufficient (D)

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