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QJEEM22Ma4aad: JEE Main 2022 Mathematics - Class 11 JEEM22Ma4aad Solution: The number of 7-digit numbers which are multiples of 11 and are …

The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____________.

Step-by-Step Verified Solution

Digits are 1,2,3,4,5,7,91,2,3,4,5,7,9

Multiple of 11→11 \rightarrow Difference of sum at even and odd place is divisible by 11 .

Let number of the form abcdefg

amp;∴(a+c+e+g)−(b+d+f)=11xamp;a+b+c+d+e+f=31amp;∴ either a+c+e+g=21 or 10amp;∴b+d+f=10 or 21\begin{aligned}&\therefore(\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g})-(\mathrm{b}+\mathrm{d}+\mathrm{f})=11 \mathrm{x} \\\\&\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=31 \\\\&\therefore \text { either } \mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=21 \text { or } 10 \\\\&\therefore \mathrm{b}+\mathrm{d}+\mathrm{f}=10 \text { or } 21\end{aligned}

Case-1

amp;a+c+e+g=21amp;b+d+f=10amp;(b,d,f)∈{(1,2,7)(2,3,5)(1,4,5)}amp;(a,c,e,g)∈{(1,4,7,9),(3,4,5,9),(2,3,7,9)}\begin{aligned}&a+c+e+g=21 \\\\&b+d+f=10 \\\\&(b, d, f) \in\{(1,2,7)(2,3,5)(1,4,5)\} \\\\&(a, c, e, g) \in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\}\end{aligned}

Case-2

amp;a+c+e+g=10amp; b+d+f=21amp;(a,b,e,g)∈{1,2,3,4)}amp;(b,d,f)&{(5,7,9)}amp;∴ Total number in case 2=3!×4!=144amp;∴ Total numbers =144+432=576\begin{aligned}&\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=10 \\\\&\mathrm{~b}+\mathrm{d}+\mathrm{f}=21 \\\\&(\mathrm{a}, \mathrm{b}, \mathrm{e}, \mathrm{g}) \in\{1,2,3,4)\} \\\\&(\mathrm{b}, \mathrm{d}, \mathrm{f}) \&\{(5,7,9)\} \\\\&\therefore \text { Total number in case } 2=3 ! \times 4 !=144 \\\\&\therefore \text { Total numbers }=144+432=576\end{aligned}

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