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BrentGMATPrepNow
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N is a 3-digit number, and the sum o 11 Jun 2018, 07:01
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Difficulty:

75% (hard)

Question Stats:

56% (02:28) correct 44% (02:41) wrong based on 194 sessions

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N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.
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B
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N is a 3-digit number, and the sum o 11 Jun 2018, 07:34
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Solution



Given:
    • N is a 3-digit number
    • Sum of the digits of N is 10
    • When N is reversed, the new number is 99 less than N

To find:
    • The value of N

Approach and Working:
If we assume N = abc, then reverse of N = cba

As N – reverse of N = 99, we can write,
    • abc – cba = 99
    Or, 100a + 10b + c – 100c – 10b – a = 99
    Or, 99 (a – c) = 99
    Or, a – c = 1

As both a and c are single-digit integers, the possible values of (a, c) = (2, 1), (3, 2), (4, 3), (5, 4)
[(6, 5) onwards values are not possible as sum of the digits is equal to 10]

Using the relation, a + b + c = 10, the possible value sets of (a, b, c) = (2, 7, 1), (3, 5, 2), (4, 3, 3), (5, 1, 4)

Analysing Statement 1
    • As per the information given in statement 1, a = 1 + c
      o This is an information that we have already derived, and not giving us any new information

Hence, statement 1 is not sufficient to answer

Analysing Statement 2
    • As per the information given in statement 2, b = 2 + a
    • From the possible values of (a, b, c) we can have only one case where it is getting satisfied = (3, 5, 2)
Therefore, the number N = 352

Hence, statement 2 is sufficient to answer

Hence, the correct answer is option B.

Answer: B
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N is a 3-digit number, and the sum o 11 Jun 2018, 07:35
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GMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
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Let's represent N as abc where:

a = hundreds digit
b = tens digit
c = ones digit

Therefore algebraically:

N = 100a + 10b + c

Reverse N = 100c + 10b + a

Given: N = Reverse N + 99
----> 100a + 10b + c = 100c + 10b + a + 99

Simplify: 99a = 99c + 99 -----> a = c + 1

From given info we know:
a = c + 1
a + b + c = 10

Our goal is to solve for N by finding a,b and c

Statement 1
Tells us a = c + 1. We already know this info so its not helpful
-------> INSUFFICIENT

Statement 2 says b = a + 2

Therefore everything can be written in terms of a:
a = a
b = a + 2
c = a - 1

Since a + b + c = 10
----> a + (a+2) + (a - 1) = 10

----> a = 3
----> c = 2
----> b = 5

Therefore N is 352.
-------------------------> sufficient

Answer: B





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N is a 3-digit number, and the sum o 11 Jun 2018, 09:25
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GMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
Show more

Since we get a 3-digit number(reversed) after subtracting 99 from N(3-digit integer), therefore N must be greater than 200.

Property:- When the digits of a 3-digit number are reversed after subtracting the original number by 99, the hundred's place digit is one more than the unit's place digit.

Question stem:- N=?

Statement1:- As mentioned above in the property, this statement doesn't provide additional info. So, st1 is insufficient.

Statement-2

The affixed table may be referred.

From the table, N=352.

Hence, statement2 is sufficient.

Ans. B

Hope my logic is correct.
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File comment: I stopped checking at 100th digit=3 because any positive integer thereafter yields sum of digits>10.
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N is a 3-digit number, and the sum o 11 Jun 2018, 14:24
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GMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
Show more

We are given the following information in the question stem:
1. Let N(3 digit number) be xyz where x - hundred's digit, y - ten's digit, z - one's digit.
2. x+y+z = 10.
3. xyz - zyx = 99(Here zyx is the reverse of the number N)

We have been asked the value of N.

1. When the hundred's digit of N is 1 more than it's units digit, there are 4 possibilities: 271,352,433,514.
Reversing these 4 numbers we will get 172,253,334,415 & each of these 4 numbers when subtracted from
the number it reverses, gives us 99. We can't find a unique value for N. (Insufficient)

2. If the tens digit of N is 2 more than its hundreds digit, there are 3 possibilities: 136,244,352. Reversing
these 3 numbers, we will get 631,442,253. Only 352 - 253 gives 99, making 352 our number (Sufficient - Option B)
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N is a 3-digit number, and the sum o 13 Jun 2018, 08:29
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GMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
Show more

Target question: What is the value of N?

Given: N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N.
Let N = wxy, where w, x and y represent the 3 digits in N
So, the first equation we can write is: w + x + y = 10

Now let's examine the situation where we REVERSE the digits.
First of all, the VALUE of N (wxy) is equal to 100w + 10x + y
Next, the VALUE of the REVERSED number (yxw) is equal to 100y + 10x + w
Since the REVERSED number is 99 less than N, we can write: 100y + 10x + w = 100w + 10x + y - 99
Subtract w from both sides: 100y + 10x = 99w + 10x + y - 99
Subtract 10x from both sides: 100y = 99w + y - 99
Subtract y from both sides: 99y = 99w - 99
Divide both sides by 99 to get: y = w - 1

Now onto the statements....

Statement 1: The hundreds digit of N is 1 more than its units digit.
In other words, w = y + 1, which is the SAME as y = w - 1
Notice that this provides NO NEW INFORMATION, since we already concluded that y = w - 1
Since statement 1 provides no new information, it is NOT SUFFICIENT

Statement 2: The tens digit of N is 2 more than its hundreds digit.
In other words, x = w + 2
Now that we've written x and w in terms of w (i.e., x = w + 2 and y = w - 1), we can take the given information (w + x + y = 10), and replace w and x
When we do so, we get: w + (w + 2) + (w - 1) = 10
Simplify: 3w + 1 = 10
Solve: w = 3
If w = 3, then x = 5, and y = 2
So, N =352
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

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Brent
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N is a 3-digit number, and the sum o 11 Sep 2020, 13:48
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BrentGMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
Show more

From the question stem, we can build up an equationfor N = 100a+10b +c. and a+b+c= 10.
100a+10b+c - 100c -10b -a = 99
99a - 99c = 99
a -c = 1

1) the same info as the stem. not sufficient.

2) b = a + 2. So, we can say, a + a+ 2 + a - 1 = 10. or, a = 3. b = 5 and c = 2. sufficient.

B is the answer.
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N is a 3-digit number, and the sum o 31 Jul 2024, 23:22
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BrentGMATPrepNow
N is a 3-digit number, and the sum of its digits is 10. When the digits of N are reversed, the new number is 99 less than N. What is the value of N?

(1) The hundreds digit of N is 1 more than its units digit.
(2) The tens digit of N is 2 more than its hundreds digit.

*kudos for all correct solutions
Show more
­Let the hunderds, tenth and unit digits of the number N be a , b, c

a + b + c =10 ------------------ (1)
N = 100a + 10 b + c

By inversing the digits the number N* = 100c + 10 b + a

From the question we have 100c + 10b + a = 100a + 10b +c + 99
This can be simplified to c = a+1---------------------(2)
Statement (1) The hundreds digit of N is 1 more than its units digit. i.e c = a+1
As we already have this equation, statement 1 is not sufficient

Statement (2) b = 2+a ------------- (3)
By solving equations (1),(2),(3) we can get the value of a,b,c
Statement (2) is sufficient.

Hence the answer is option (B).
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