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vagabundo [1.1K]
4 years ago
12

Hi, What is the lowest positive integer greater than 1, which when divided by 5 or 8 leaves a remainder of 1

Mathematics
1 answer:
ludmilkaskok [199]4 years ago
3 0
Very interesting question!

Let's try to get something that would work for just 5 for now.
Any of these numbers would leave a remainder of 1 when divided by just 5:
6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, ...
Notice that they all look like 5*(something) + 1. Basically, they are all multiples of 5 plus 1.

Let's see what would work for just 8:
9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, ...

Are there any matches... AHA! It's 81. This is the lowest positive integer greater than 1 that follows the rule for both 5 and 8.
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Answer:

see the attachments for the two solutions

Step-by-step explanation:

When the given angle is opposite the shorter of the given sides, there will generally be two solutions. The exception is the case where the triangle is a right triangle (the ratio of the given sides is equal to the sine of the given angle). If the given angle is opposite the longer of the given sides, there is only one solution.

When a side and its opposite angle are given, as here, the law of sines can be used to solve the triangle(s). When the given angle is included between two given sides, the law of cosines can be used to solve the (one) triangle.

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Here, the law of sines can be used to solve the triangle:

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