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ravi1522
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How many integers from 0 to 50, inclusive, have a remainder 15 May 2024, 08:50
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­Lets write few number which will have remainder 1 when divided by 3 
1,4,7,10,13,........,49 
so this series in AP so  to get total number nth term =first term +(N-1)d
49=1+(N-1)*3
Hence total number will be 17.
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How many integers from 0 to 50, inclusive, have a remainder 02 Sep 2024, 23:21
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Is this considered a word problem?
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Bunuel
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How many integers from 0 to 50, inclusive, have a remainder 03 Sep 2024, 00:13
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Matthewjd24
Is this considered a word problem?
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No, this would be considered a question on remainders.
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How many integers from 0 to 50, inclusive, have a remainder 29 Sep 2024, 00:25
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I used a little bit of AP formula here-

As per question, we get x=3q+1
So, the set x has (1,4,7,....,49). We want the total number count in this set. If you notice, this looks like an AP, with common difference (d)= 3 and first term (a1) as 1.

Now, use the formula, 49= 1 + (n-1)3
=> n= 17

Ans. C.

Hope it helps.
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How many integers from 0 to 50, inclusive, have a remainder 22 Nov 2025, 12:13
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X=3B+1
X can be the following: 1,4,7,10...

Since I know that 1 has to be the first number included, I did (50-1/3) + 1. I got 17.66. Only up to 17 numbers work.

Bunuel, can you let me know if this works?
Bunuel
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A. 15
B. 16
C. 17
D. 18
E. 19

Algebraic way:

Integer have a remainder of 1 when divided by 3 implies \(n=3p+1\), where \(p\) is an integer \(\geq{0}\), so \(n\) can take the following values: 1, 4, 7, ...

\(n=3p+1\leq{50}\);

\(3p\leq{49}\);

\(p\leq{16\frac{1}{3}}\)

Hence, \(p\), can take 17 values from 0 to 16, inclusive.

Answer: C.­
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