# 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, 11 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 $$142 \times 7 + 3 < 1000$$)

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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.

- 4 years, 11 months ago

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 $$7k+3$$.

- 4 years, 11 months ago

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.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

143

- 4 years, 11 months ago

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.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

thanks

- 4 years, 11 months ago

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

- 4 years, 11 months ago