Question

find the largest positive integer n such that n3 + 100 is divisible by n+10? plz explain it...

Answer 1

Heya User,

---> For [ n + 10 ] to divide [ n³ + 100 ], let's follow the steps :->

--> Firstly, --> a | b, means that a completely divides b without leaving any remainder...

Now, --> [ n + 10 ] | [ n³ + 100 ]
======> [ n + 10 ] | [ n³ + 1000 ] - 900 ---> ( a + b ) | ( a³ + b³ )
======> [ n + 10 ] | -900

Hence, biggest no. for which [ n + 10 ] divides -900 is :->
---> [ n + 10 ] = 900
---> n = 890..

VERIFICATION :->
---> [ 890 + 10 ] = 900
---> [ 890 ]³ + 100 = 704969000 + 100 = 704969100

Clearly, 900 * 783299 = 704969100..

Hence, your req.d answer is 890...

Answer 2

The largest positive integer n is, 890.

What is positive integer example?

• An integer is positive if it is greater than zero. Example 1, 2, 3 . . .
• Negative Integers: An integer is negative if it is less than zero. Example -1, -2, -3 .

What is positive integer number?

• Positive integers are all the whole numbers greater than zero: 1, 2, 3, 4, 5, ... .
• Negative integers are all the opposites of these whole numbers: -1, -2, -3, -4, -5, …. We do not consider zero to be a positive or negative number.

What is negative and positive integers?

• Whole numbers, figures that do not have fractions or decimals, are also called integers.
• They can have one of two values, positive or negative.
• Positive integers have values greater than zero.
• Negative integers have values less than zero.
• Zero is neither positive nor negative.

According to the question:

By division we find that, $n^{3}+100=(n+10)\left(n^{2}-10 n+100\right)-900$.

Therefore, if n+10 divides $n^{3}$+100.

Then it must also divide 900.

Since we are looking for largest n, n is maximized whenever n+10.

Since the largest divisor of 900 is 900.

We must have, n+10=900.

n=890

Hence, The largest positive integer n is, 890.

Learn more about positive integer here,

https://brainly.in/question/97157?msp_poc_exp=5

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