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 $3$ ?

#NumberTheory #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

A Brilliant Member - 7 years, 7 months ago

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

Matt McNabb - 7 years, 7 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.

Aiman Rafeed - 7 years, 7 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 $7k+3$ .

Tanishq Aggarwal - 7 years, 7 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.

Aiman Rafeed - 7 years, 7 months ago

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

A Former Brilliant Member - 7 years, 7 months ago

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

Daniel Liu - 7 years, 7 months ago

143

ANSHUL AGARWAL - 7 years, 7 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.

ANSHUL AGARWAL - 7 years, 7 months ago

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

Suraj Sonule - 7 years, 7 months ago

@Suraj Sonule – thanks

ANSHUL AGARWAL - 7 years, 7 months ago

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

A Brilliant Member - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$