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24 Jul 2023, 08:00
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B
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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
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02 Feb 2024, 01:56
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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
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:
Thus:
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.
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:
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
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 \(\frac{ last - first }{ multiple}+1=\frac{945-105}{42}+1=9\)
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
General Discussion
24 Jul 2023, 08:34
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answer is B
first 3 digit multiple of 7 = 105
last 3 digit multiple of 7 = 994
No of multiples of 7 = ((994-105)/7) + 1 = 127 + 1 = 128
we need to find those multiples of 7 which are NOT multiples of 6 and 15. so subtract multiples of 42 and 105. (LCM of 7,6 and 7,15 respectively) But we need to add multiples of LCM of 7,6,15. since they would have been subtracted twice.
Following similar steps as above like we did for 7; no of multiples of 42 = 21, no of multiples of 105 = 9, no of multiples of 210 = 4
So answer is 128 - 21 - 9 + 4 = 102. answer is B.
first 3 digit multiple of 7 = 105
last 3 digit multiple of 7 = 994
No of multiples of 7 = ((994-105)/7) + 1 = 127 + 1 = 128
we need to find those multiples of 7 which are NOT multiples of 6 and 15. so subtract multiples of 42 and 105. (LCM of 7,6 and 7,15 respectively) But we need to add multiples of LCM of 7,6,15. since they would have been subtracted twice.
Following similar steps as above like we did for 7; no of multiples of 42 = 21, no of multiples of 105 = 9, no of multiples of 210 = 4
So answer is 128 - 21 - 9 + 4 = 102. answer is B.
