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20 May 2019, 07:26
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Difficulty:
85%
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Question Stats:
47% (02:34) correct
53%
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wrong
based on 112
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Consider a set T={x, x+1, x+2, x+3, x+4, x+5}, where x can take any integral value between 1 and 999 (both are inclusive). How many of these sets don't contain any multiple of 7?
A. 141
B. 142
C. 143
D. 144
E. 145
A. 141
B. 142
C. 143
D. 144
E. 145
20 May 2019, 08:27
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If x is one greater than some multiple of 7, then when we divide x by 7, the remainder will be 1, when we divide x+1 by 7, the remainder will be 2, and so on. Dividing all six numbers in our set by 7 will give us, in order, remainders of 1, 2, 3, 4, 5 and 6. So we won't have any multiples of 7 in our set in that case, but in any other case we will.
So the question is just asking, "how many positive integers less than 1000 give a remainder of 1 when you divide them by 7?" We can list the numbers we're talking about:
1, 8, 15, 22, .... , 981, 988, 995
and now to count how many things are in this list, since it is equally spaced, we can use the formula n = (range/space) + 1, so n = (995-1)/7 + 1 = 994/7 + 1 = 142 + 1 = 143. Or if we want to avoid formulas, we can just add 6 to everything in the list:
7, 14, 21, ...., 994, 1001
and divide everything by 7
1, 2, 3, ...., 142, 143
and we have a list just as long as our first list, which clearly has 143 things in it.
So the question is just asking, "how many positive integers less than 1000 give a remainder of 1 when you divide them by 7?" We can list the numbers we're talking about:
1, 8, 15, 22, .... , 981, 988, 995
and now to count how many things are in this list, since it is equally spaced, we can use the formula n = (range/space) + 1, so n = (995-1)/7 + 1 = 994/7 + 1 = 142 + 1 = 143. Or if we want to avoid formulas, we can just add 6 to everything in the list:
7, 14, 21, ...., 994, 1001
and divide everything by 7
1, 2, 3, ...., 142, 143
and we have a list just as long as our first list, which clearly has 143 things in it.
19 Oct 2022, 17:34
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Can’t do any better than the answer already posted.
Following the pattern is another way to see how the sets of consecutive integers behave.
[1, 2, 3, 4, 5, 6]
[2, 3, 4, 5, 6, 7]
…..
[7, 8, 9, 10, 11, 12]
When X is equal to one of the 7 consecutive positive integers [1 through 7], you will only have one set with NO multiple of 7.
So we can’t count 1 for each positive multiple of 7 up to 999
7 (142) = 700 + 280 + 14 = 994
We will have 142 sets that do not have a multiple of 7.
Then one more set:
[995, 996, 997, 998, 999, 1000]
142 + 1 =
143
Posted from my mobile device
Following the pattern is another way to see how the sets of consecutive integers behave.
[1, 2, 3, 4, 5, 6]
[2, 3, 4, 5, 6, 7]
…..
[7, 8, 9, 10, 11, 12]
When X is equal to one of the 7 consecutive positive integers [1 through 7], you will only have one set with NO multiple of 7.
So we can’t count 1 for each positive multiple of 7 up to 999
7 (142) = 700 + 280 + 14 = 994
We will have 142 sets that do not have a multiple of 7.
Then one more set:
[995, 996, 997, 998, 999, 1000]
142 + 1 =
143
Posted from my mobile device
