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Both a and b are perfect squares, and the product a×b is divisible by 10 as well as 15. By which of the following the product a×b may NOT be divisible?
- a)60
- b)50
- c)120
- d)150
- e)225
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Both a and b are perfect squares, and the product a×b is divisib...
Given that both a and b are perfect squares and their product, a * b, is divisible by both 10 and 15, we can conclude that a * b must also be divisible by the highest common factor (HCF) of 10 and 15. Additionally, since a and b are perfect squares, their product, a * b, must also be a perfect square.
We can express a * b as (5 * 2 * 3)2 * (Integer)^2, where the common factors of 10 and 15, i.e., 5 and 3, are squared. Therefore, a * b can be written as 900 * I2, where I represents any integer.
Considering the given options, the only number that may not divide the product a * b is 120.
Hence, the answer is option C.
Most Upvoted Answer
Both a and b are perfect squares, and the product a×b is divisib...
Given that both a and b are perfect squares and their product, a * b, is divisible by both 10 and 15, we can conclude that a * b must also be divisible by the highest common factor (HCF) of 10 and 15. Additionally, since a and b are perfect squares, their product, a * b, must also be a perfect square.
We can express a * b as (5 * 2 * 3)2 * (Integer)^2, where the common factors of 10 and 15, i.e., 5 and 3, are squared. Therefore, a * b can be written as 900 * I2, where I represents any integer.
Considering the given options, the only number that may not divide the product a * b is 120.
Hence, the answer is option C.
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Both a and b are perfect squares, and the product a×b is divisib...
Understanding the Problem
To determine which product \( a \times b \) may NOT be divisible by, we first need to analyze the conditions given.
Given Conditions
- Both \( a \) and \( b \) are perfect squares.
- \( a \times b \) is divisible by both 10 and 15.
Divisibility Analysis
1. Divisibility by 10:
- \( 10 = 2 \times 5 \)
- Therefore, \( a \times b \) must have at least one factor of 2 and one factor of 5.
2. Divisibility by 15:
- \( 15 = 3 \times 5 \)
- Thus, \( a \times b \) must also have at least one factor of 3.
Combining Conditions
From the above, \( a \times b \) needs:
- At least one factor of \( 2 \)
- At least one factor of \( 3 \)
- At least two factors of \( 5 \) (since both \( a \) and \( b \) are perfect squares, the exponent of each prime factor must be even in a perfect square).
Checking Each Option
- 60: \( 60 = 2^2 \times 3 \times 5 \) (Divisible)
- 50: \( 50 = 2 \times 5^2 \) (Divisible)
- 120: \( 120 = 2^3 \times 3 \times 5 \) (Divisible)
- 150: \( 150 = 2 \times 3 \times 5^2 \) (Divisible)
- 225: \( 225 = 3^2 \times 5^2 \) (Divisible)
Why 120 is NOT Possible
- Since \( a \) and \( b \) are perfect squares, \( a \times b \) must maintain even exponents across all prime factors.
- The factorization \( 120 = 2^3 \times 3 \times 5 \) includes an odd exponent for 2 and 3, which violates the perfect square condition.
Thus, \( a \times b \) may NOT be divisible by 120.
To determine which product \( a \times b \) may NOT be divisible by, we first need to analyze the conditions given.
Given Conditions
- Both \( a \) and \( b \) are perfect squares.
- \( a \times b \) is divisible by both 10 and 15.
Divisibility Analysis
1. Divisibility by 10:
- \( 10 = 2 \times 5 \)
- Therefore, \( a \times b \) must have at least one factor of 2 and one factor of 5.
2. Divisibility by 15:
- \( 15 = 3 \times 5 \)
- Thus, \( a \times b \) must also have at least one factor of 3.
Combining Conditions
From the above, \( a \times b \) needs:
- At least one factor of \( 2 \)
- At least one factor of \( 3 \)
- At least two factors of \( 5 \) (since both \( a \) and \( b \) are perfect squares, the exponent of each prime factor must be even in a perfect square).
Checking Each Option
- 60: \( 60 = 2^2 \times 3 \times 5 \) (Divisible)
- 50: \( 50 = 2 \times 5^2 \) (Divisible)
- 120: \( 120 = 2^3 \times 3 \times 5 \) (Divisible)
- 150: \( 150 = 2 \times 3 \times 5^2 \) (Divisible)
- 225: \( 225 = 3^2 \times 5^2 \) (Divisible)
Why 120 is NOT Possible
- Since \( a \) and \( b \) are perfect squares, \( a \times b \) must maintain even exponents across all prime factors.
- The factorization \( 120 = 2^3 \times 3 \times 5 \) includes an odd exponent for 2 and 3, which violates the perfect square condition.
Thus, \( a \times b \) may NOT be divisible by 120.
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