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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
3 years, 10 months 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 · 3 years, 10 months ago

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hint:all such number will form an A.P wih comon diffrence 7. Varad Kausadikar · 3 years, 10 months 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 · 3 years, 10 months ago

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@Aiman Rafeed 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 · 3 years, 10 months ago

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@Tanishq Aggarwal 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 · 3 years, 10 months ago

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Try it for small numbers (say 1-30) and look for a pattern. Matt McNabb · 3 years, 10 months 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 · 3 years, 10 months ago

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143 Anshul Agarwal · 3 years, 10 months ago

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@Anshul Agarwal 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 · 3 years, 10 months ago

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@Anshul Agarwal good job......solved problem in classical and simplest way unlike dat pappe Suraj Sonule · 3 years, 10 months ago

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@Suraj Sonule thanks Anshul Agarwal · 3 years, 9 months ago

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@Anshul Agarwal Explain how. This is not a question. Someone has posted this since he/she needs help in this. Paramjit Singh · 3 years, 10 months ago

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