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26 May 2020, 08:56
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C
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Difficulty:
95%
(hard)
Question Stats:
33% (02:25) correct
67%
(02:21)
wrong
based on 87
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If a subset of integers is chosen from between 1 to 3000, inclusive, such that no two integers add up to a multiple of nine, then what can be the maximum number of elements in the subset? Include both 1 and 3000?
(A) 1332
(B) 1333
(C) 1336
(D) 1668
(E) 1672
Are You Up For the Challenge: 700 Level Questions
(A) 1332
(B) 1333
(C) 1336
(D) 1668
(E) 1672
Are You Up For the Challenge: 700 Level Questions
26 May 2020, 10:26
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IMO C.
Let's see what integers can be included in the set:
1, 2,3,4, (5,6,7,8,9,10,11,12,13,14,15,16,17,18…..
Notice that thee underlined numbers cannot be included since the sum of any one of these underlined numbers and a particular non-underlined number will result in a multiple of 9.
So, after every 4 numbers the next 5 numbers cannot be considered. This cycle of 4 possible and next 5 not possible integers continues 333 times (till 2997).
The next three numbers 2998, 2999 and 3000 can also be included in the set.
We get a total of (333 X 4) + 3 = 1335 numbers.
Now, notice that number 9 can also be included in this set (since 9 plus any other number from this set will not be a multiple of 9).
Therefore, 1335+1 = 1336 numbers.
Answer C.
Let's see what integers can be included in the set:
1, 2,3,4, (5,6,7,8,9,10,11,12,13,14,15,16,17,18…..
Notice that thee underlined numbers cannot be included since the sum of any one of these underlined numbers and a particular non-underlined number will result in a multiple of 9.
So, after every 4 numbers the next 5 numbers cannot be considered. This cycle of 4 possible and next 5 not possible integers continues 333 times (till 2997).
The next three numbers 2998, 2999 and 3000 can also be included in the set.
We get a total of (333 X 4) + 3 = 1335 numbers.
Now, notice that number 9 can also be included in this set (since 9 plus any other number from this set will not be a multiple of 9).
Therefore, 1335+1 = 1336 numbers.
Answer C.
General Discussion
26 May 2020, 10:33
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The sum of the two integers can't be a multiple of 9.
Sum pairs of 9: (0,9) , (1,8) , (2,7) , (3,6) , (4,5)
From the above we understand that for:
(0,9) - if we take a multiple of 9, we can't take any other multiple of 9. This pair gives us 1 number (it can be any multiple of 9). Count = 1
(1,8) - if we take a number such that the remainder is 1 when divided by 9, then we need to discard all those numbers which leave a remainder of 8. Count = 334
Similarly,
(2,7) Count = 334
(3,6) Count = 334
(4,5) Count = 333, since the 334th occurrence will be 3001 (which is >3000).
Total = 1+333 + 334*3 = 334*4 = 1336
Sum pairs of 9: (0,9) , (1,8) , (2,7) , (3,6) , (4,5)
From the above we understand that for:
(0,9) - if we take a multiple of 9, we can't take any other multiple of 9. This pair gives us 1 number (it can be any multiple of 9). Count = 1
(1,8) - if we take a number such that the remainder is 1 when divided by 9, then we need to discard all those numbers which leave a remainder of 8. Count = 334
Similarly,
(2,7) Count = 334
(3,6) Count = 334
(4,5) Count = 333, since the 334th occurrence will be 3001 (which is >3000).
Total = 1+333 + 334*3 = 334*4 = 1336
