Question

Find the sum of all 3 digit natural numbers which are multiples of 11 and 13

Answer 1

if seperately ...three digit natural nos that are multiples of 11 is
110,121,132,143,......,990
so here d= 11
FIRST term 110
last term 990
so 990=110+(n-1)11
or n=81
so total sum here is (81/2)*(2*110+(80*11))=44550
again multiple of 13 is 104,117,130,143,.....,988
so 988=104+(n-1)*13
or n=69
so here total sum 69/2 (2*104+(69-1)*13)=37674
so total = 44550+37674=82224

Answer 2

Conclusion

Natural Number.

Natural numbers are those in mathematics that are used for counting and ordering. Cardinal numbers are those that are used for counting, and ordinal numbers are those that are used for ordering.

Main solution

If seperately, There digit natural number that are multiples of 11 is

110, 1211, 132, 143,.......,990

So, Here $d=11$

First Term 110

$So, 900 = 100+(n-1)11\\or,n=81$

So, Total sum here is $(\frac{81}{2}) \times (2\times110+(80\times11)) = 44550$

Again, multiple of

$13 is 104,117,130,143,...........,988\\So, 988=104=(n-1)*13\\or, n=69$

so here total sum

$69/2 (2\times 104 + (69-1)\times13)=37674\\So, total = 44550 + 37647 = 82224$

To learn more about Natural Number.

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