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Need Help In Number Theory

If we divide all the numbers from \(1\) to \(1000\) by \(7\).For how many numbers will the remainder be \(3\) ?

Note by Fazla Rabbi
4 years, 1 month ago

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

Paramjit Singh - 4 years, 1 month ago

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

Varad Kausadikar - 4 years ago

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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.

Aiman Rafeed - 4 years ago

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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 \(7k+3\).

Tanishq Aggarwal - 4 years ago

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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.

Aiman Rafeed - 4 years ago

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

Matt McNabb - 4 years ago

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

Daniel Liu - 4 years ago

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143

Anshul Agarwal - 4 years, 1 month ago

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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.

Anshul Agarwal - 4 years ago

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

Suraj Sonule - 4 years ago

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@Suraj Sonule thanks

Anshul Agarwal - 4 years ago

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

Paramjit Singh - 4 years, 1 month ago

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