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02 Feb 2024, 01:52
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How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15?
A. 98
B. 102
C. 106
D. 110
E. 114
A. 98
B. 102
C. 106
D. 110
E. 114
02 Feb 2024, 01:52
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Official Solution:
How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15?
A. 98
B. 102
C. 106
D. 110
E. 114
Let's determine the number of multiples of 7 from 100 to 1000, inclusive. The number of multiples of an integer within a range can be calculated using the following formula:
• \(\frac{\text{last multiple in the range - first multiple in the range} }{\text{multiple} }+1\)
Thus:
• The number of multiples of 7 in the given range is \(\frac{ last - first}{multiple}+1=\frac{994 -105}{7}+1=128\)
Since we need only those multiples that are not also multiples of 6 or 15, we should subtract from that number the multiples of 42, which is the least common multiple of 7 and 6, and the multiples of 105, which is the least common multiple of 7 and 15.
• The number of multiples of 42 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{966-126}{42}+1=21\)
• The number of multiples of 105 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{945-105}{105}+1=9\)
However, both counts above include the multiples of 6, 7, and 15, namely, the multiples of 210. Hence, we need to subtract that number from 21 + 9 = 30 to avoid double-counting those:
• The number of multiples of 210 in the given range is \(\frac{ last - first}{multiple}+1=\frac{840 -210}{210}+1=4\)
Therefore, the number of three-digit positive integers that are multiples of 7 but not multiples of 6 or 15 is \(128 - (21 + 9 - 4) = 102\).
Answer: B
How many three-digit positive integers are multiples of 7 but not multiples of 6 or 15?
A. 98
B. 102
C. 106
D. 110
E. 114
Let's determine the number of multiples of 7 from 100 to 1000, inclusive. The number of multiples of an integer within a range can be calculated using the following formula:
• \(\frac{\text{last multiple in the range - first multiple in the range} }{\text{multiple} }+1\)
Thus:
• The number of multiples of 7 in the given range is \(\frac{ last - first}{multiple}+1=\frac{994 -105}{7}+1=128\)
Since we need only those multiples that are not also multiples of 6 or 15, we should subtract from that number the multiples of 42, which is the least common multiple of 7 and 6, and the multiples of 105, which is the least common multiple of 7 and 15.
• The number of multiples of 42 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{966-126}{42}+1=21\)
• The number of multiples of 105 in the given range is \(\frac{ last - first }{ multiple}+1=\frac{945-105}{105}+1=9\)
However, both counts above include the multiples of 6, 7, and 15, namely, the multiples of 210. Hence, we need to subtract that number from 21 + 9 = 30 to avoid double-counting those:
• The number of multiples of 210 in the given range is \(\frac{ last - first}{multiple}+1=\frac{840 -210}{210}+1=4\)
Therefore, the number of three-digit positive integers that are multiples of 7 but not multiples of 6 or 15 is \(128 - (21 + 9 - 4) = 102\).
Answer: B
07 Jul 2024, 14:27
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Dear Bunuel,
Thank yoou for the explanation. Is there a fast way to find the first and last multiples in a range for numbers such as 42 and 105?
Thank yoou for the explanation. Is there a fast way to find the first and last multiples in a range for numbers such as 42 and 105?
