Need Help In Number Theory

If we divide all the numbers from 11 to 10001000 by 77.For how many numbers will the remainder be 33 ?

Note by Fazla Rabbi
5 years, 12 months ago

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Hint: All such numbers will be of form 7k+37k+3 with 0k100070 \leq k \leq \lfloor \frac{1000}{7}\rfloor (As 142×7+3<1000142 \times 7 + 3 < 1000)

A Brilliant Member - 5 years, 12 months ago

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

Matt McNabb - 5 years, 12 months ago

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Note that every integer of the form 7k+37k+3 is a number that gives a remainder 33 when divided 77. ...(1)...(1)

Equally importantly, whenever an integer divided by 77 gives a remainder of 33, it must be of the form 7k+37k+3. ...(2)...(2)

Let xx be such an integer.

So, 1x10001\leq x \leq 1000 17k+31000\Leftrightarrow 1 \leq 7k+3 \leq 1000 [ For some integer k] 27k997 \Leftrightarrow -2 \leq 7k \leq 997 27k10007 \Leftrightarrow \frac {-2}{7} \leq k \leq \frac {1000}{7}

There are 143143 integer solutions to this inequality from 00 to 142142. From statement (1)(1), we can say, for each kk we can have a corresponding x satisfying the condition in the problem and from statement (2)(2), we can say, there is no other xx's other than these which will satisfy the condition. So there are 143143 integers from 11 to 10001000 which give a remainder of 33 when divided by 77.

Hope this somewhat helps.

Aiman Rafeed - 5 years, 12 months 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+37k+3.

Tanishq Aggarwal - 5 years, 12 months 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 - 5 years, 12 months ago

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

A Former Brilliant Member - 5 years, 12 months ago

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Different hint: Consider 11000(mod7)1\rightarrow 1000\pmod{7}. How many numbers 1x10001\le x\le 1000 are there such that x(mod7)x\equiv \pmod{7}?

Daniel Liu - 5 years, 12 months ago

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143

ANSHUL AGARWAL - 5 years, 12 months 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 - 5 years, 12 months ago

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

Suraj Sonule - 5 years, 12 months ago

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@Suraj Sonule thanks

ANSHUL AGARWAL - 5 years, 11 months ago

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

A Brilliant Member - 5 years, 12 months ago

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