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Given that 674pq0 is divisible by 3 and 11 For this question we will use divisibility rule 3 divisibility rule = sum of digits should be divisible by 3. 11 divisibility rule = sum of even places - sum of odd places = 0 ,11,22,33...... Here we have , 674pq0 According to divisibility rule of 3 => 6+7+4+p+q+0 = 17+p+q => it should be multiple of 3 So, the possible value of p+q = 1 , 4 , 7, 10, 13..... Now you can see the option or you can apply 11 rule as well. Then from the option only option 2 and 3 are satisfying the condition that sum should be 1,4,7..... but option 2 is not satisfying the divisibility rule of 11 . So option 2 can not be the answer of this question . Option 3 which is p = 5 and q =2 is satisfying both the rules of 3 and 11.
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