JEE Advanced 2018 · Paper 1 · Mathematics Q9

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MathematicsNumerical+3Question 45 of 54

Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \ldots$ , and $Y$ be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \ldots$ . Then, the number of elements in the set $X \cup Y$ is _____.

Your answer

Answer and worked solution

Official answer: 3748

Common terms. $X$ has terms $\equiv1\pmod5$ , up to $1+5\cdot2017=10086$ . $Y$ has terms $\equiv2\pmod7$ , from 9 up to $9+7\cdot2017=14128$ .

The smallest number that is both $\equiv1\pmod5$ and $\equiv2\pmod7$ is 16 (and $16\ge9$ ). The common terms form the AP $16,51,86,\ldots$ with difference 35, up to 10086:
$\left\lfloor\frac{10086-16}{35}\right\rfloor+1=287+1=288.$

Union.
$|X\cup Y|=2018+2018-288=3748.$

Answer: 3748

Topics:Arithmetic progression Sets and operations

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JEE Advanced 2018 · Paper 1, Mathematics question 9