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{1, 5, 9, 13, …125} is series with 4 as common difference.

Lets find pairs which has sum 146 : { 21, 125} , { 25, 121} , { 29, 117} ,..........{ 69 ,77 }. // 13 set possible

These are total 13 pairs, and remaining {1,5,9,13,} {73 also Because 73+ 73 makes 146 but here no repeatation.}i.e. 6 numbers.

Apply pigeonhole principle : take any Number from 13 set + these 6 Numbers = 19 numbers

After that if you pick(2nd number) any number you get atleast two numbers whose sum is 146, So, answer is 19 + 1= 20

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Another approach:A more general one though here since d sum is 146 which is a small value. Since the difference is 4 therefore all d numbers in d set can be obtained. Starting with 1+4=5,5+4=9,9+4=13,13+4=17,17+4=21,21+4=25,25+4=29,29+4=31,...and so on until d last value i.e. 125. Either by adding from front or subtracting from back all d no's. Like in my case I"ve summed up. When all the member no's of set r obtained then do d following:

The nos obtained r following:{1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81,85,89,93,97,101,105,109,113,117,121,125}

From 146 start subtracting with d last no. Like 146-125=21, which is gain a member of d set. In a similar manner 146-121=25, both 21 and 25 r members of d set. Though it hardly took any time. Hence d following pairs were obtained:

(125,21),(121,25),(113,33),(117,29),(109,37),(105,41),(101,45),(97,49),(93,53),(57,89),(85,61),(81,65),(77,69). These pairs total upto 146 which r 13 in no. If repetition not allowed, earlier mentioned excluding (73,73). 

Final asn is 13. This approach is used for a smaller set. 

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