If 27y is a multiple of 9 where y is a digit . What is the minimum value of Y.

Mathematics

1 answer:

stiv31 [10]2 years ago

4 0

<h3>Answer: 0</h3>

Explanation:

The single digits are {0,1,2,3,4,5,6,7,8,9}

So naturally 0 is the first thing to test.

The number 270 is a multiple of 9 because 9*30 = 270. The digits of 270 add to 2+7+0 = 9 showing that it's a multiple of 9. Refer to the "divisibility by 9 rule".

The next possible value of y will be 9 units higher than y = 0, so we go for y = 9. Meaning that 279 is also a multiple of 9 (because 2+7+9 = 18 is a multiple of 9).

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igor_vitrenko [27]

The value of $a_{5}$ is 2500, when $a_{1}=4$ and $5a_{n-1}$ .

Given that, $a_{1}=4$ and $5a_{n-1}$ .

We need to find the value of $a_{5}$ .

<h3>What is an arithmetic sequence?</h3>

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Now, to find the value of $a_{5}$ :

$a_{2} =5a_{2-1}=5a_{1}=5 \times4=20$