Question: Consider 3 consecutive prime integers. Twice the first integer is 5 more than the third integer and second is 4 less than the third

Consider 3 consecutive prime integers. Twice the first integer is 5 more than the third integer and second is 4 less than the third integer. Sum of the integers is 41. Find the smallest integer among the three. 1. 11 13 3. 7 5. None of these 2. 4.
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Solution 1 Let first second and third integer be a b and ... View full answer
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