JEE Main 2021 — Mathematics Algebra

The prime factorisation of the given number is $2040={2}^{3}\times 3\times 5\times 17$2040=23×3×5×17

The H.C.F. of $n$n and $2040$2040 is $1$1 if $n$n is not a multiple of $2,3,5$2,3,5 and $17.$17.

Hence $n$n cannot be multiple of $2,3,5$2,3,5 and $17$17

Then sum is $n(1)-(n(2)+n(3)+n(5)+n(17)-n(6)-n(10)-n(34)-n(15)-n(51)-n(85))+n(30))$n(1)−(n(2)+n(3)+n(5)+n(17)−n(6)−n(10)−n(34)−n(15)−n(51)−n(85))+n(30))

Where $n(a)$n(a) means the sum of all numbers belonging to the set ${1,2,\ldots \ldots .100}$1,2,…….100 which are divisible by $a$a

$$=\frac{100\times 101}{2}-(\frac{2\times 50\times 51}{2}+\frac{3\times 33\times 34}{2}+\frac{5\times 20\times 21}{2}+\frac{17\times 5\times 6}{2}-\frac{6\times 16\times 17}{2}-\frac{10\times 10\times 11}{2}-\frac{34\times 2\times 3}{2}-\frac{15\times 6\times 7}{2}-51-85+180)$$=2100×101​−(22×50×51​+23×33×34​+25×20×21​+217×5×6​−26×16×17​−210×10×11​−234×2×3​−215×6×7​−51−85+180)

$$=5050-2550-1683-1050-255+816+550+102+315+51+85-180$$=5050−2550−1683−1050−255+816+550+102+315+51+85−180

$$=1251$$=1251

Alternate Solution:

Thus, the sum of all the required numbers taking the common elements only once, is $(1+3+5+\ldots ..+99)-(3+9+15+21+\ldots \ldots +99)-(5+25+35+55+65+85+95)-(17)$(1+3+5+…..+99)−(3+9+15+21+……+99)−(5+25+35+55+65+85+95)−(17)

Now, using the sum of $n$n terms of an arithmetic progression i.e. ${S}_{n}=\frac{n}{2}[a+l],$Sn​=2n​[a+l], where $a$a is the first term and $l$l is the last term.

$$=\frac{50}{2}[1+99]-\frac{17}{2}[3+99]-(365)-17$$=250​[1+99]−217​[3+99]−(365)−17

$$=\frac{50}{2}(100)-\frac{17}{2}(102)-365-17$$=250​(100)−217​(102)−365−17

$$=2500-867-365-17$$=2500−867−365−17

$$=1251.$$=1251.