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Note that every integer of the form $7k+3$ is a number that gives a remainder $3$ when divided $7$. $...(1)$

Equally importantly, whenever an integer divided by $7$ gives a remainder of $3$, it must be of the form $7k+3$. $...(2)$

Let $x$ be such an integer.

So, $1\leq x \leq 1000$$\Leftrightarrow 1 \leq 7k+3 \leq 1000$[ For some integer k]$\Leftrightarrow -2 \leq 7k \leq 997$$\Leftrightarrow \frac {-2}{7} \leq k \leq \frac {1000}{7}$

There are $143$ integer solutions to this inequality from $0$ to $142$. From statement $(1)$, we can say, for each $k$ we can have a corresponding x satisfying the condition in the problem and from statement $(2)$, we can say, there is no other $x$'s other than these which will satisfy the condition. So there are $143$ integers from $1$ to $1000$ which give a remainder of $3$ when divided by $7$.

We know that no. should be of the form (7x+3). First no.should be 3 and like this last no. should be 994+3 = 997 . That makes an AP with a =3, last term L =997 and common difference d =7. So, L =a + (n-1)d , where n is the no. of terms
997 =3 + (n-1)* 7,
994= (n-1)* 7,
(n-1)= 142,
n = 143.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestHint: All such numbers will be of form $7k+3$ with $0 \leq k \leq \lfloor \frac{1000}{7}\rfloor$ (As $142 \times 7 + 3 < 1000$)

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Try it for small numbers (say 1-30) and look for a pattern.

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Note that every integer of the form $7k+3$ is a number that gives a remainder $3$ when divided $7$. $...(1)$

Equally importantly, whenever an integer divided by $7$ gives a remainder of $3$, it must be of the form $7k+3$. $...(2)$

Let $x$ be such an integer.

So, $1\leq x \leq 1000$ $\Leftrightarrow 1 \leq 7k+3 \leq 1000$

[ For some integer k]$\Leftrightarrow -2 \leq 7k \leq 997$ $\Leftrightarrow \frac {-2}{7} \leq k \leq \frac {1000}{7}$There are $143$ integer solutions to this inequality from $0$ to $142$. From statement $(1)$, we can say, for each $k$ we can have a corresponding x satisfying the condition in the problem and from statement $(2)$, we can say, there is no other $x$'s other than these which will satisfy the condition. So there are $143$ integers from $1$ to $1000$ which give a remainder of $3$ when divided by $7$.

Hope this somewhat helps.

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You could condense your first two statements by saying

an integer divided by 7 gives a remainder of 3 if and only if the number is of the form $7k+3$.Log in to reply

Right, but the meaning of

"if and only if"sometimes isn't clear to many people. They confuse it with only"if". That's why I did it.Log in to reply

hint:all such number will form an A.P wih comon diffrence 7.

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Different hint: Consider $1\rightarrow 1000\pmod{7}$. How many numbers $1\le x\le 1000$ are there such that $x\equiv \pmod{7}$?

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143

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We know that no. should be of the form (7x+3). First no.should be 3 and like this last no. should be 994+3 = 997 . That makes an AP with a =3, last term L =997 and common difference d =7. So, L =a + (n-1)d , where n is the no. of terms

997 =3 + (n-1)* 7, 994= (n-1)* 7, (n-1)= 142, n = 143.

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good job......solved problem in classical and simplest way unlike dat pappe

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Explain how. This is not a question. Someone has posted this since he/she needs help in this.

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