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In the first 1000 natural numbers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?
11
14
12
13
Step 1:The first sequence: Numbers leaving a remainder of 4 when divided by 7: 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, .... The second sequence: Numbers leaving a remainder of 9 when divided by 11: 9, 20, 31, 42, 53, 64, ..... From the listing of the two sequences we can identify the first number that is a part of the both the sequences is 53. Step 2:The common difference of the first sequence is 7 and that of the second sequence is 11. Elements common to both the sequences will have a common difference that is the LCM of 7 and 11. 77 is the LCM of 7 and 11. Every 77th number after 53 will be a term common to both the series. So, the terms that are common to both the arithmetic sequences can be expressed as 77k + 53. Because we are interested in the first 1000 natural numbers, k will take values from 0 to 12. i.e., a total of 13 values.
By: Amit Kumar ProfileResourcesReport error
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