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For any integral of n, 32n + 9n + 5 when divided by 3 will leave the remainder
- a)1
- b)2
- c)0
- d)5
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
For any integral of n, 32n+ 9n + 5 when divided by 3 will leave the re...
32n + 9n + 5 = 3 × 32n-1 + 9n + 3 + 2
= 3(32n-1 + 3n + 1) + 2
As, 3(32n-1 + 3n + 1) is completely divisible by 3.
Therefore,
Remainder is 2
= 3(32n-1 + 3n + 1) + 2
As, 3(32n-1 + 3n + 1) is completely divisible by 3.
Therefore,
Remainder is 2
Most Upvoted Answer
For any integral of n, 32n+ 9n + 5 when divided by 3 will leave the re...
Explanation:
To find the remainder when (32n + 9n + 5) is divided by 3, we need to use the property that the remainder when a number is divided by 3 is the same as the remainder when the sum of its digits is divided by 3.
Let's simplify the expression (32n + 9n + 5) to (41n + 5).
Now, let's find the remainder when the sum of the digits of (41n + 5) is divided by 3.
When we add the digits of 41n, we get:
4 + 1 + n = 5 + n
So, the sum of the digits of (41n + 5) is:
5 + n + 5 = 10 + n
Now, we need to find the remainder when (10 + n) is divided by 3.
We can write (10 + n) as (9 + 1 + n).
When (9 + 1 + n) is divided by 3, we get a remainder of 1, because:
9 is divisible by 3 with no remainder.
1 is not divisible by 3, so it leaves a remainder of 1.
n may or may not be divisible by 3, but it doesn't matter because any multiple of 3 added to 1 will still leave a remainder of 1 when divided by 3.
So, the remainder when (41n + 5) is divided by 3 is 1.
Therefore, the correct answer is option B) 2.
To find the remainder when (32n + 9n + 5) is divided by 3, we need to use the property that the remainder when a number is divided by 3 is the same as the remainder when the sum of its digits is divided by 3.
Let's simplify the expression (32n + 9n + 5) to (41n + 5).
Now, let's find the remainder when the sum of the digits of (41n + 5) is divided by 3.
When we add the digits of 41n, we get:
4 + 1 + n = 5 + n
So, the sum of the digits of (41n + 5) is:
5 + n + 5 = 10 + n
Now, we need to find the remainder when (10 + n) is divided by 3.
We can write (10 + n) as (9 + 1 + n).
When (9 + 1 + n) is divided by 3, we get a remainder of 1, because:
9 is divisible by 3 with no remainder.
1 is not divisible by 3, so it leaves a remainder of 1.
n may or may not be divisible by 3, but it doesn't matter because any multiple of 3 added to 1 will still leave a remainder of 1 when divided by 3.
So, the remainder when (41n + 5) is divided by 3 is 1.
Therefore, the correct answer is option B) 2.
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For any integral of n, 32n+ 9n + 5 when divided by 3 will leave the remaindera)1b)2c)0d)5Correct answer is option 'B'. Can you explain this answer?
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