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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24? 04 Oct 2023, 17:45
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D
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63% (02:25) correct 37% (02:36) wrong based on 272 sessions

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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24?

(1) Units' digit of 2^m is a prime number
(2) Units' digit of 3^m is the smallest odd prime number
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D
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GMAT Focus 1: 735 Q90 V89 DI81
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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24? 04 Oct 2023, 21:42
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RenB
Given that m is an integer, Is (m) (m + 1)(m + 2) divisible by 24?
(1) Units' digit of 2^m is a prime number
(2) Units' digit of 3^m is the smallest odd prime number
Show more


When should be m(m+1)(n+2), that is product of three consecutive numbers be multiple of 8?
a) m is even.
We can be sure that the product is multiple of 3.
We can also be sure that the product is multiple of 8, as m and m+2 are even and one of them is multiple of 4 for sure .

b) m is odd.
Again, the product is multiple of 3.
However, the product will be multiple of 8, only when m+1 is multiple of 8.

(1) Unit digit of \(2^m\) is a prime number
Unit digit of consecutive powers of 2 are 2,4,8,6,2,4…..
Here, digit 2 is prime and repeats at powers of 1, 5…..4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient

(2) Units' digit of 3^m is the smallest odd prime number
Unit digit of consecutive powers of 3 are 3,9,7,1,3,9…..
Here, digit 3 is smallest odd prime and repeats at powers of 1, 5…..4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient


D
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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24? 01 Nov 2024, 00:07
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chetan2u
RenB
Given that m is an integer, Is (m) (m + 1)(m + 2) divisible by 24?
(1) Units' digit of 2^m is a prime number
(2) Units' digit of 3^m is the smallest odd prime number


When should be m(m+1)(n+2), that is product of three consecutive numbers be multiple of 8?
a) m is even.
We can be sure that the product is multiple of 3.
We can also be sure that the product is multiple of 8, as m and m+2 are even and one of them is multiple of 4 for sure .

b) m is odd.
Again, the product is multiple of 3.
However, the product will be multiple of 8, only when m+1 is multiple of 8.

(1) Unit digit of \(2^m\) is a prime number
Unit digit of consecutive powers of 2 are 2,4,8,6,2,4.....
Here, digit 2 is prime and repeats at powers of 1, 5.....4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient

(2) Units' digit of 3^m is the smallest odd prime number
Unit digit of consecutive powers of 3 are 3,9,7,1,3,9.....
Here, digit 3 is smallest odd prime and repeats at powers of 1, 5.....4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient


D
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chetan2u thank you for providing the solution above. Can you please explain how the answer will be D when we can take 1 as the value of m given it will satisfy both the equations? If we take m=1, then, the equation will not be divisible by 24 whereas when we take m=5 (the next possible value), the equation will be divisible by 24. Should the answer not be E in this case?
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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24? 01 Nov 2024, 01:08
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UdhavK
chetan2u
RenB
Given that m is an integer, Is (m) (m + 1)(m + 2) divisible by 24?
(1) Units' digit of 2^m is a prime number
(2) Units' digit of 3^m is the smallest odd prime number


When should be m(m+1)(n+2), that is product of three consecutive numbers be multiple of 8?
a) m is even.
We can be sure that the product is multiple of 3.
We can also be sure that the product is multiple of 8, as m and m+2 are even and one of them is multiple of 4 for sure .

b) m is odd.
Again, the product is multiple of 3.
However, the product will be multiple of 8, only when m+1 is multiple of 8.

(1) Unit digit of \(2^m\) is a prime number
Unit digit of consecutive powers of 2 are 2,4,8,6,2,4.....
Here, digit 2 is prime and repeats at powers of 1, 5.....4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient

(2) Units' digit of 3^m is the smallest odd prime number
Unit digit of consecutive powers of 3 are 3,9,7,1,3,9.....
Here, digit 3 is smallest odd prime and repeats at powers of 1, 5.....4k+1
Substitute m as 4k+1 => \((4k+1)(4k+2)(4k+3)=(4k+1)*2(2k+1)*(4k+3)=(odd)*(2*odd)*(odd)\). Thus the product will be multiple of 2 but not 4 or 8.
Answer is no as m is multiple of 6 but not of 24.
Sufficient


D
chetan2u thank you for providing the solution above. Can you please explain how the answer will be D when we can take 1 as the value of m given it will satisfy both the equations? If we take m=1, then, the equation will not be divisible by 24 whereas when we take m=5 (the next possible value), the equation will be divisible by 24. Should the answer not be E in this case?
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When m = 5, (m)(m + 1)(m + 2) = 5 * 6 * 7, which is not divisible by 24.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:


  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this and similar questions.

Hope it helps.
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Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24? 18 Aug 2025, 10:50
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Question rephrased: Does the numerator have 24 = (2^3)*(3^1) "in stock?"

Pre-work
i) The numerator contains 3 sequential numbers: (Even)(Odd)(Even) or (Odd)(Even)(Odd)
ii) 3's occur every 3rd number {1,2,3 4,5,6 7,8,9...}
Therefore, the numerator will always have at least (2^1) and a (3^1) in stock

iii) We only need to know whether the numerator has (2^2) in stock

Question simplified: Is m odd?


(1)
i) For 2^#, the pattern of units digits repeats every 4 numbers (e.g. 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16)
ii) The only prime number resulting from 2^# occurs when 2^1 = 2 = prime
Therefore, m = 1 = odd
Sufficient [ELIMINATE BCE]

(2)
i) For 3^#, the pattern of units digits repeats every 4 numbers (e.g. 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81)
The prime numbers among 3's pattern of unit digits are 3 and 7 (1 is not prime)
Therefore, the smallest odd prime is 3, which makes m = 1, (1+4=5), (5+4=9), ... = ODD
Sufficient [ELIMINATE A]

Answer: D
RenB
Given that m is an integer, Is (m)(m + 1)(m + 2) divisible by 24?

(1) Units' digit of 2^m is a prime number
(2) Units' digit of 3^m is the smallest odd prime number
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