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The sum of all two digit numbers each of which leaves remainder 3 when divided by 5 is

Please solve it without without using arithmetic progression. I have solved it by arithmetic progression and got the answer but I have to solve it by some other method?

Thomas Andrews
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Ankit Kumar
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    I don't see any obvious way without using that the set of numbers are numbers of the form $5k+3$ for $k=2$ to $19.$ Using $j=k-2$, this is $5j+13$ for $j=0$ to $17.$ Then you have the sum is $$\sum_{k=0}^{17}(5k+13)=18\cdot 13 + 5\cdot \frac{17\cdot 18}{2}$$ – Thomas Andrews Jun 14 '19 at 17:48
  • It is unclear what you mean by solving it "using arithmetic progressions." Can you show us your solution if you are seeking another? – Thomas Andrews Jun 14 '19 at 17:56
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    Yeah sure, first of all I supposed that let the number be 5q+3 then I found that after putting q = 2 I got 13 as my first number and 18 as second and 23 as third number and also 96 as my greatest two digit number which follow this rule. After that I found that it's an arithmetic progression as a = 13 and common difference = 5 and we know that the sum of n term of arithmetic progression is n÷2(a+l) by using this I got my sum as 999. – Ankit Kumar Jun 14 '19 at 18:00
  • But this question is of chapter real numbers in my book so I have to use the formulas including the chapter real numbers only. So I thought of asking this question as I am not getting any way out from real numbers. – Ankit Kumar Jun 14 '19 at 18:01
  • We don't know what is in the chapter, so we can't say what is in there. – Thomas Andrews Jun 14 '19 at 18:07

3 Answers3

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We can use a shortcut to calculate the sum:

  • We know the first term in the list (13) and the last term (98). There are 18 terms in total (= (98-13)/5 + 1)
  • If you pair up the first term and the last term, and the second term and the second-to-last-term, etc. the pairs all have the same sum, 111.
  • So we have 9 pairs of numbers that sum to 111. Their total is 999.
user326210
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The sum is $$\sum _2^{19} ( 5k+3) = 5\sum _2^{19} ( k) + 3(18) = 5(189)+54 =999$$

Note that $$\sum _2^{19} k =\sum _1^{19} k -1 = \frac {19(20)}{2} -1=189$$

Mohammad Riazi-Kermani
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The first number is 13 and the last number is 98. 13+98 = 111.

The second number is 18 and the second last number is 93. 18+93 = 111.

How many 111's will there be?

B. Goddard
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