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 soon#### Find 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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