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A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals
- a)31
- b)63
- c)75
- d)91
- e)87
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A positive whole numberMless than 100 is represented in base 2 notatio...
Since in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5
So the no. may be 31 or 91
Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1
So the no. may be 31 or 91
Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1
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A positive whole numberMless than 100 is represented in base 2 notatio...
To solve this problem, let's analyze the given information step by step:
1. The last digit in all three representations (base 2, base 3, and base 5) is 1.
- This means that the number M can be expressed as M = 2a + 1, M = 3b + 1, and M = 5c + 1, where a, b, and c are positive integers.
2. Exactly two out of the three representations have a leading digit of 1.
- This means that M is not divisible by 2, 3, or 5, except for the last digit.
Let's consider the possible values for M:
- If M is divisible by 2, then it can be expressed as M = 2k, where k is a positive integer.
- In this case, the last digit of M in base 2 is 0, which contradicts the given information.
- Therefore, M is not divisible by 2.
- If M is divisible by 3, then it can be expressed as M = 3k, where k is a positive integer.
- In this case, the last digit of M in base 3 is 0, which contradicts the given information.
- Therefore, M is not divisible by 3.
- If M is divisible by 5, then it can be expressed as M = 5k, where k is a positive integer.
- In this case, the last digit of M in base 5 is 0, which contradicts the given information.
- Therefore, M is not divisible by 5.
Now, let's consider the possible values for M based on the above analysis:
- M cannot be divisible by 2, 3, or 5.
- The smallest number that satisfies this condition is 1, but it doesn't satisfy the condition of being less than 100.
- The next number that satisfies this condition is 7, which can be expressed as 7 = 2(3) + 1, 7 = 3(2) + 1, and 7 = 5(1) + 1.
- However, 7 is not less than 100.
Hence, the only remaining possibility is that M = 91, which satisfies all the given conditions:
- 91 = 2(45) + 1
- 91 = 3(30) + 1
- 91 = 5(18) + 1
Therefore, the correct answer is option D) 91.
1. The last digit in all three representations (base 2, base 3, and base 5) is 1.
- This means that the number M can be expressed as M = 2a + 1, M = 3b + 1, and M = 5c + 1, where a, b, and c are positive integers.
2. Exactly two out of the three representations have a leading digit of 1.
- This means that M is not divisible by 2, 3, or 5, except for the last digit.
Let's consider the possible values for M:
- If M is divisible by 2, then it can be expressed as M = 2k, where k is a positive integer.
- In this case, the last digit of M in base 2 is 0, which contradicts the given information.
- Therefore, M is not divisible by 2.
- If M is divisible by 3, then it can be expressed as M = 3k, where k is a positive integer.
- In this case, the last digit of M in base 3 is 0, which contradicts the given information.
- Therefore, M is not divisible by 3.
- If M is divisible by 5, then it can be expressed as M = 5k, where k is a positive integer.
- In this case, the last digit of M in base 5 is 0, which contradicts the given information.
- Therefore, M is not divisible by 5.
Now, let's consider the possible values for M based on the above analysis:
- M cannot be divisible by 2, 3, or 5.
- The smallest number that satisfies this condition is 1, but it doesn't satisfy the condition of being less than 100.
- The next number that satisfies this condition is 7, which can be expressed as 7 = 2(3) + 1, 7 = 3(2) + 1, and 7 = 5(1) + 1.
- However, 7 is not less than 100.
Hence, the only remaining possibility is that M = 91, which satisfies all the given conditions:
- 91 = 2(45) + 1
- 91 = 3(30) + 1
- 91 = 5(18) + 1
Therefore, the correct answer is option D) 91.
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A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer?.
A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer? for CAT 2025 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for CAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A positive whole numberMless than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. ThenMequalsa)31b)63c)75d)91e)87Correct answer is option 'D'. Can you explain this answer?.
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