Find the sum of positive integers less than 150 but greater than 30 that are divisible by 7.
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Answer:
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Answer:
(First: How many multiples of 7 are there between 20 and 150?)
There are 19 of them. Because there are 21 = floor[150/7] = floor [21.143..] positive multiples of 7 less than 150. Two of these, 7 and 14, are less than 20. So 21 - 2 = 19.
Second: Keep in mind that if we factor 7 from the sum of T consecutive multiples of 7, the result will be 7 times the sum of the T corresponding consecutive integer quotients.
Third: Remember: (sum i = 1 to N; ( i ) ) = N*(N-1)/2 // pseudo C programming language notation
(sum i = a, b; ( i )} = { (sum i = 1,b; ( i )) - (sum i = 1, a; ( i )) }
= b*(b-1)/2- a*(a-1)/2 .
Finally: the
answer = 7*{sum i = 3,19; ( i ) } = 7*{3+4+5+ .. +18+19}
= 7*{sum i = a, b; ( i ) } // where a=3 and b=19
= 7*{ (sum i = 1, 19; ( i )) - (sum i = 1,3); ( i )) }
= 7*((b*(b-1))/2 - a*(a-1))/2)
= 7*(19*18/2 - 3*2/2) = 7*(19*9 - 3) = 7*(171 - 3) = 7*168
= 1176 .
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