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Just wanted to add that I didn't know any way how stmt 2 could give an answer, but just in case there could be a possibility, and since i had no logic to check, i took random numbers satisfying stmt 2, in this case 6 and 8, and substituted in the polynomial expression, both gave different remainders, so i was sure that stmt 2 has to be insufficient.

Guys, this is a very interesting question , I have seen this type for the first time. I have couple of questions

1) if remainder for t/7 is known we can know the remainder for t^2/7 , and vice verca is not possible ? ie if the the remainder for t^2/7 is known t/7 is not known ?

2) where can I get additional information on these types of questions ?

Guys, this is a very interesting question , I have seen this type for the first time. I have couple of questions

1) if remainder for t/7 is known we can know the remainder for t^2/7 , and vice verca is not possible ? ie if the the remainder for t^2/7 is known t/7 is not known ?

2) where can I get additional information on these types of questions ?

Help !

Actually there is a remainder rule, that works only for additions, subtractions and multiplications, but not for divisions.

Ill illustrate this rule with an example.

25/7 - R= 4 41/7 - R= 6 addition (25+41)/7 should give remainder of 4+6, but 10 is greater than 7, so the final remainder will be that of 10/7, ie 3 checking this rule - 25+41 = 66..... 66/7 - R = 3

subtraction (41-25)/7 should give remainder 6-4 = 2...... check: 41-25 = 16. .....16/7 - R= 2

multiplication 41*25/7 should give remainder of (6*4)/7, ie 24/7 -> R= 3 check: 41*25 = 1025....1025/7 -> R= 3

Now square is nothing but (t/7) * (t/7), so applying multiplication rule, we get that remainder should be R*R

Square root is division, and therefore the rule doesnt apply to square roots.