The given question is an easy to moderate level difficulty question combining concepts in Arithmetic Progression and elementary number theory - LCM in particular.
Question: 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
- 10
Explanatory Answer
Video explanation will be added soonFind the first number common to both the sequences.
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.
Compute common difference of terms common to both sequences and find number of terms
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 satisfy both the conditions.
Compute common difference of terms common to both sequences and find number of terms
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 satisfy both the conditions.
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.
Choice D is the correct answer.
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