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QOTD#14 When x is divided by 3, the remainder is 1. When x
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14 Nov 2016, 15:14
14 Nov 2016, 15:14
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When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?
Drill 4
Question: 5
Page: 293
Question: 5
Page: 293
Show: :: OA
4
Re: QOTD#14 When x is divided by 3, the remainder is 1. When x
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14 Nov 2016, 15:21
14 Nov 2016, 15:21
3
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Explanation
To solve this question, write it out.
Since there are fewer numbers that yield a remainder of 2 when divided by 7, start there. The first such number is 2, and thereafter they increase by 7; the rest of the list is thus 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, and 93. Rather than list out all the numbers that yield a remainder of 1 when divided by 3, just select the numbers that meet the requirement from the list you already have: Only 16, 37, 58, and 79 do, so there are 4 values for x.
Hence the correct answer is 4.
To solve this question, write it out.
Since there are fewer numbers that yield a remainder of 2 when divided by 7, start there. The first such number is 2, and thereafter they increase by 7; the rest of the list is thus 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, and 93. Rather than list out all the numbers that yield a remainder of 1 when divided by 3, just select the numbers that meet the requirement from the list you already have: Only 16, 37, 58, and 79 do, so there are 4 values for x.
Hence the correct answer is 4.
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Re: QOTD#14 When x is divided by 3, the remainder is 1. When x
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25 Nov 2016, 17:46
25 Nov 2016, 17:46
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sandy wrote:
When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
Now onto the question....
When x is divided by 7, the remainder is 2
Possible values of x are: 2, 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93
When x is divided by 3, the remainder is 1
Possible values of x are: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97
The two sets have 4 numbers in common.
Answer:
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ASIDE: Notice that each of the shared values (16, 37, 58, and 79) are 21 greater than the previous shared value. Also notice that 21 is the least common multiple (LCM) of 3 and 7.
So, once we found 1 value in common, we could have just kept adding 21 to that value to find the subsequent values.
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