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The number of integers x such that 0.25 < />X< 200="" and="" />X + 2 is perfectly divisible by either 3 or 4, is
Correct answer is '5'. Can you explain this answer?
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The number of integers x such that 0.25 XX + 2 is perfectly divisible...
At x = 0, 2X = 1 which is in the given range [0.25,200]
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2X + 2 = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.
At x = 1, 2X = 2 which is in the given range [0.25, 200]
2x + 2 = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.
At x = 2, 2X = 4 which is in the given range [0.25, 200]
2x + 2 = 4 + 2 = 6 which is divisible by 3. Hence, x = 2 is one possible solution.
At x = 3, 2X = 8 which is in the given range [0.25, 200]
2x + 2 = 8 + 2 = 3 which is not divisible by 3 or 4. Hence, x = 3 can't be a solution.
At x = 4, 2X = 16 which is in the given range [0.25,200]
2X + 2 = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.
At x = 5, 2X = 32 which is in the given range [0.25,200]
2X + 2 = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution.
At x = 6, 2X = 64 which is in the given range [0.25,200]
2X + 2 = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.
At x = 7, 2X = 128 which is in the given range [0.25,200]
2X + 2 = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution.
At x = 8, 2X = 256 which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.
Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.
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The number of integers x such that 0.25 XX + 2 is perfectly divisible...
Explanation:
Divisibility by 3:
- For a number to be divisible by 3, the sum of its digits must be divisible by 3.
- In this case, the number is 0.25XX + 2.
- The sum of the digits of 0.25XX + 2 is 2 + 5 + X + X + 2 = 9 + 2X.
- For 9 + 2X to be divisible by 3, 2X must be divisible by 3.
- This implies X can take values 0, 3, 6, or 9.
Divisibility by 4:
- For a number to be divisible by 4, the number formed by the last two digits must be divisible by 4.
- In this case, the last two digits are X2.
- For X2 to be divisible by 4, X can take values 0, 2, 4, 6, or 8.
Common values of X:
- The common values of X that satisfy both the conditions are 0 and 6.
- Therefore, the integers x such that 0.25XX + 2 is perfectly divisible by either 3 or 4 are 0 and 6.
- Hence, the total number of integers x = 2.
Therefore, the correct answer is '2'.
Divisibility by 3:
- For a number to be divisible by 3, the sum of its digits must be divisible by 3.
- In this case, the number is 0.25XX + 2.
- The sum of the digits of 0.25XX + 2 is 2 + 5 + X + X + 2 = 9 + 2X.
- For 9 + 2X to be divisible by 3, 2X must be divisible by 3.
- This implies X can take values 0, 3, 6, or 9.
Divisibility by 4:
- For a number to be divisible by 4, the number formed by the last two digits must be divisible by 4.
- In this case, the last two digits are X2.
- For X2 to be divisible by 4, X can take values 0, 2, 4, 6, or 8.
Common values of X:
- The common values of X that satisfy both the conditions are 0 and 6.
- Therefore, the integers x such that 0.25XX + 2 is perfectly divisible by either 3 or 4 are 0 and 6.
- Hence, the total number of integers x = 2.
Therefore, the correct answer is '2'.
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The number of integers x such that 0.25 XX + 2 is perfectly divisible by either 3 or 4, isCorrect answer is '5'. Can you explain this answer?
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