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