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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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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? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about 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? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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