Math, asked by spp37193, 8 months ago

how many 3 digit number are there which leave remainder 3 on division by 1

Answers

Answered by Anonymous

0

Answer:

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Answered by rajdeep1778

0

Answer:

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Step-by-step explanation:

First number 100≡2(mod7)

101≡3(mod7) is the first number having remainder 7

Last number 999≡5(mod7)

Hence 997≡3(mod7) is last number having remainder 7.

So total numbers having remainders 7 between 100 and 999 are

(997−101)÷7+1=128+1=129

(1 added to include 101)

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