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 \(7\).

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 \(7k+3\).

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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## 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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