Question

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Que 2. Sum of 3 digit numbers divided by 5 gives the remainder of 3 ?
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Answer 1

Answer:

Smallest 3 digit natural number divisible by 3=102

Largest 3 digit natural number divisible by 3=999

∴3 digit number divisible by 3 are:

102,105,108,.....999

this is an AP with a=102 and d=3

no.of terms

999=a+(n−1)d

999=102+(n−1)3

3

897

=(n−−)⇒(n−1)=299

∴n=300

Sum of all numbers = Sum of 300 terms of AP

=

2

300

[2a+)300−1)d]

=

2

300

[2×(102)+(299)3]

=(150)[204+891]

Sum of all number of series =(150)[1101]=165150

∴ Sum of all 3 digit number divisible by 3 is 165150