what is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers and the sum of eleven consecutive integers?
2007-09-13 12:08:53 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Consider twice the sum of nine consecutive integers, starting with n:
...... n + (n+1) + (n+2)... (n+8)
(n+8) + (n+7) + (n+6)... n
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2n+8 + 2n+8 + 2n+8... 2n+8
Which is 9*(2n+8). Therefore, the sum of 9 consecutive integers starting with n is 9*(2n+8)/2. Similarly, the sum of 10 consecutive integers starting with m is 10*(2m+9)/2 and the sum of 11 consecutive integers starting with k is 11*(2m+10)/2. So any such number must be a multiple of 9, 5, and 11. The smallest such number is 495. And indeed:
51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 = 495
45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 = 495
40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 = 495
So 495 is the smallest such positive integer.
2007-09-13 12:25:00 · answer #1 · answered by Pascal 7 · 1⤊ 0⤋
Well, an integer is either a negative or a positive number or zero. So, the smallest set of nine consecutive integers is -4 to +4 and their sum is zero!
As far as ten go, I'll pick -4 through +5, which add to five.
Eleven? I'll pick -5 to +5, which again sum to zero.
2007-09-13 12:28:03 · answer #2 · answered by PMP 5 · 0⤊ 0⤋