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If the number A is even, which of the following will be true?
- a)3A will always be divisible by 6
- b)3A + 5 will always be divisible by 11
- c)(A2 + 3)/4 will be divisible by 7
- d)All of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If the number A is even, which of the following will be true?a)3A will...
To determine which statement(s) will always be true if the number A is even, let's analyze each option:
a) 3A will always be divisible by 6:
- If A is even, it can be written as A = 2k, where k is an integer.
- Substituting this value into the expression 3A: 3A = 3(2k) = 6k.
- Since 6 is a multiple of 6, we can conclude that 3A will always be divisible by 6.
- Therefore, statement a) is true.
b) 3A - 5 will always be divisible by 11:
- We know that A is even, so A = 2k, where k is an integer.
- Substituting this value into the expression 3A - 5: 3A - 5 = 3(2k) - 5 = 6k - 5.
- However, 6k - 5 is not necessarily divisible by 11.
- For example, if k = 1, then 6(1) - 5 = 1, which is not divisible by 11.
- Therefore, statement b) is not always true.
c) (A^2 + 3)/4 will be divisible by 7:
- Let's consider a few values of A to determine if the expression is always divisible by 7.
- If A = 2, then (A^2 + 3)/4 = (4 + 3)/4 = 7/4, which is not divisible by 7.
- If A = 4, then (A^2 + 3)/4 = (16 + 3)/4 = 19/4, which is not divisible by 7.
- If A = 6, then (A^2 + 3)/4 = (36 + 3)/4 = 39/4, which is not divisible by 7.
- From these examples, we can see that the expression is not always divisible by 7.
- Therefore, statement c) is not always true.
Based on our analysis, only statement a) is always true.
a) 3A will always be divisible by 6:
- If A is even, it can be written as A = 2k, where k is an integer.
- Substituting this value into the expression 3A: 3A = 3(2k) = 6k.
- Since 6 is a multiple of 6, we can conclude that 3A will always be divisible by 6.
- Therefore, statement a) is true.
b) 3A - 5 will always be divisible by 11:
- We know that A is even, so A = 2k, where k is an integer.
- Substituting this value into the expression 3A - 5: 3A - 5 = 3(2k) - 5 = 6k - 5.
- However, 6k - 5 is not necessarily divisible by 11.
- For example, if k = 1, then 6(1) - 5 = 1, which is not divisible by 11.
- Therefore, statement b) is not always true.
c) (A^2 + 3)/4 will be divisible by 7:
- Let's consider a few values of A to determine if the expression is always divisible by 7.
- If A = 2, then (A^2 + 3)/4 = (4 + 3)/4 = 7/4, which is not divisible by 7.
- If A = 4, then (A^2 + 3)/4 = (16 + 3)/4 = 19/4, which is not divisible by 7.
- If A = 6, then (A^2 + 3)/4 = (36 + 3)/4 = 39/4, which is not divisible by 7.
- From these examples, we can see that the expression is not always divisible by 7.
- Therefore, statement c) is not always true.
Based on our analysis, only statement a) is always true.
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Community Answer
If the number A is even, which of the following will be true?a)3A will...
Only the first option can be verified to be true in this case.
If A is even, 3A would always be divisible by 6 as it would be divisible by both 2 and 3.
Options b and c can be seen to be incorrect by assuming the value of A as 4.
If A is even, 3A would always be divisible by 6 as it would be divisible by both 2 and 3.
Options b and c can be seen to be incorrect by assuming the value of A as 4.
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If the number A is even, which of the following will be true?a)3A will always be divisible by 6b)3A + 5 will always be divisible by 11c)(A2 + 3)/4 will be divisible by 7d)All of theseCorrect answer is option 'A'. Can you explain this answer?
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