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17 Jan 2021, 08:58
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Find the least number of digits in the number formed with only zeroes and ones such that the number is divisible by 225.
A. 4
B. 5
C. 9
D. 11
E. 24
A. 4
B. 5
C. 9
D. 11
E. 24
17 Jan 2021, 10:25
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DisciplinedPrep
Now 225=25*9.
So the number has to have 2 Zeores in it and
To be divisible by 9, the digits should add up to 9. As only 1s are allowed, we should have 9 ones.
Thus 9 ones and 2 zeroes => 9+2=11 digits
Number is 11111111100
D
General Discussion
ashik1720002
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10 Sep 2024, 13:53
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To find the least number of digits in a binary number (a number formed with only zeroes and ones) that is divisible by 225, we need to ensure the number is divisible by both 9 and 25, since
225=9×25.
Divisibility by 25
For a number to be divisible by 25, it must end in either 00, 25, 50, or 75. Since the number can only use the digits 0 and 1, the last two digits of the number must be 00.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. Since we can only use 0 and 1, the sum of the digits of our number is equal to the count of ones in the number. Therefore, the number must contain a number of ones that is divisible by 9.
Combining the Conditions
The number must:
End in 00 to be divisible by 25.
Have a total number of ones divisible by 9 to be divisible by 9.
Let's determine the smallest number of digits needed:
Minimum Length Calculation:
For the number to end in 00, we must include at least two trailing zeros.
The sum of digits (count of ones) needs to be divisible by 9.
Let's start with the smallest possible number of ones divisible by 9, which is 9. Therefore, the number needs to have 9 ones and 2 trailing zeros.
The total number of digits = Number of ones + Number of trailing zeros = 9 (ones) + 2 (trailing zeros) = 11 digits.
Verification:
We will verify if a binary number with 11 digits (9 ones and 2 trailing zeros) is the smallest such number that meets the criteria. The simplest arrangement is having 9 ones followed by 2 zeros at the end:
Example number: 11111111100
Check:
Divisibility by 25: Ends in 00 (satisfies this condition).
Sum of digits: 9 (which is divisible by 9).
Thus, the smallest binary number with 11 digits that meets both conditions is indeed possible, so the answer is:
D. 11
225=9×25.
Divisibility by 25
For a number to be divisible by 25, it must end in either 00, 25, 50, or 75. Since the number can only use the digits 0 and 1, the last two digits of the number must be 00.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. Since we can only use 0 and 1, the sum of the digits of our number is equal to the count of ones in the number. Therefore, the number must contain a number of ones that is divisible by 9.
Combining the Conditions
The number must:
End in 00 to be divisible by 25.
Have a total number of ones divisible by 9 to be divisible by 9.
Let's determine the smallest number of digits needed:
Minimum Length Calculation:
For the number to end in 00, we must include at least two trailing zeros.
The sum of digits (count of ones) needs to be divisible by 9.
Let's start with the smallest possible number of ones divisible by 9, which is 9. Therefore, the number needs to have 9 ones and 2 trailing zeros.
The total number of digits = Number of ones + Number of trailing zeros = 9 (ones) + 2 (trailing zeros) = 11 digits.
Verification:
We will verify if a binary number with 11 digits (9 ones and 2 trailing zeros) is the smallest such number that meets the criteria. The simplest arrangement is having 9 ones followed by 2 zeros at the end:
Example number: 11111111100
Check:
Divisibility by 25: Ends in 00 (satisfies this condition).
Sum of digits: 9 (which is divisible by 9).
Thus, the smallest binary number with 11 digits that meets both conditions is indeed possible, so the answer is:
D. 11
