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06 Mar 2007, 05:02
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How many sets of two or more consecutive integers are there such that no element is negative and the sum of the elements is 210?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
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06 Mar 2007, 06:09
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(A), which is 4
I do not have any formula but 210 has to be divisible by odd numbers.
For ex.
210/3 = 70, so 69+70+71 = 210.
similarly, 210/5 = 52, so 50+51+52+53+54 = 210
But if it is divisible by even number such as 2, then summation of 2 consecutive numbers can not be 210.
I do not have any formula but 210 has to be divisible by odd numbers.
For ex.
210/3 = 70, so 69+70+71 = 210.
similarly, 210/5 = 52, so 50+51+52+53+54 = 210
But if it is divisible by even number such as 2, then summation of 2 consecutive numbers can not be 210.
06 Mar 2007, 07:40
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answer is 8 (D).
for sequences with odd count, rdg's method is very good:
the middle number is always the average, so you need to find a all odd divisors of 210 - these are 3,5,7,15,21 (and 105... see below)
and you get:
3: 210/3 = 70 => 69+70+71=210
5: 210/5 = 42 => 40+41+42+43+44=210
7: 210/7 = 30 => 27+28+29+30+31+32+33=210
15: 210/15 = 14 => 7+8+...+14+..+21=210
21: 210/21 = 10 => 0+1+2+...+10+...+20=210
for 105 this doesn't work, as the resulting sequence will involve negative numbers
now we try to find sequences with even count: such sequences always have a k+0.5 average where k is an integer. so if 2n is te number of terms in the sequence, then 2n(k+0.5)=210 => n(2k+1)=210 since (2k+1) is odd, then n must be even. so the length of the sequences can be 2n, where n is an even divisor of 210 these are 2,6,10,14,20...
n=2 (length is 4) => k= 52 => 51+52+53+54=210
n=6 (length is 12) => k= 17 => 12+13+14+15+16+17+..+23=210
n=10 (length is 20) => k=10 => 1+2+3+..10+11+...20=210
for longer sequences you eventually need negative numbers...
all in all, we found 8 sequences of non-negative consecutive integers whose sum is 210.
for sequences with odd count, rdg's method is very good:
the middle number is always the average, so you need to find a all odd divisors of 210 - these are 3,5,7,15,21 (and 105... see below)
and you get:
3: 210/3 = 70 => 69+70+71=210
5: 210/5 = 42 => 40+41+42+43+44=210
7: 210/7 = 30 => 27+28+29+30+31+32+33=210
15: 210/15 = 14 => 7+8+...+14+..+21=210
21: 210/21 = 10 => 0+1+2+...+10+...+20=210
for 105 this doesn't work, as the resulting sequence will involve negative numbers
now we try to find sequences with even count: such sequences always have a k+0.5 average where k is an integer. so if 2n is te number of terms in the sequence, then 2n(k+0.5)=210 => n(2k+1)=210 since (2k+1) is odd, then n must be even. so the length of the sequences can be 2n, where n is an even divisor of 210 these are 2,6,10,14,20...
n=2 (length is 4) => k= 52 => 51+52+53+54=210
n=6 (length is 12) => k= 17 => 12+13+14+15+16+17+..+23=210
n=10 (length is 20) => k=10 => 1+2+3+..10+11+...20=210
for longer sequences you eventually need negative numbers...
all in all, we found 8 sequences of non-negative consecutive integers whose sum is 210.
