hi! thanks for the answer but I think there is an error. They have asked for non-prime integers and 7,13,19 are prime numbers

Using the dividend divisor relationship, that is, dividend = (Divisor× Quotient) + Remainder, let the positive integer which on dividing by 6 leaves remainder 1 or 2 be n which will be represented as;

𝑛 = \(6 \ell _1 + 1\) and 𝑛 = \(6 \ell _2 + 2\) [\(\ell_1\) and \(\ell_2\) are integer quotient]

𝑛 =\(6 \ell _1 + 1\)

Considering different values of \(\ell_ 1\): 0, 1, 2, 3, …

𝑛 = 1, 7, 13, 19, …

And, 𝑛 = \(6 \ell _2 + 2\) − −−→ 2,8,14 …

Hence, the non-prime integers that lie between 0 and 20 which leaves either 1 or 2 as remainder when divided by 6 will be 1, 8, 14

Ans. is 3

Since we need remainder as 1 or 2, the number has to be more than 6 and 1 or 2 more than a multiple pf 6. Non prime integers between 0-20 but more than 6: 8,9,10,12,14,15,16,18,20. Of these, select numbers 1 or 2 more than multiple a of 6 : 8, 14, 20
Answer: 3

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