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There are 27 different three-digit integers that can be formed using o
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02 Feb 2022, 06:40
02 Feb 2022, 06:40
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There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
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Re: There are 27 different three-digit integers that can be formed using o
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02 Feb 2022, 06:51
02 Feb 2022, 06:51
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Carcass wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
This is a great example of how the GRE often rewards students for thinking outside the box.
Here, we can apply a divisibility rule that says integer N is divisible by 3 if and only if the sum of the digits of N is divisible by 3.
For example, we know that 11112 is divisible by 3, because 1+1+1+1+5=9, and 9 is divisible by 3.
Notice that 1+2+3 = 6, and 6 is divisible by 3
This means any 3-digit integer consisting of a 1, a 2 and a 3 must be divisible by 3
If each of the 27 integers in the sum is divisible by 3, then the sum must also be divisible by 3.
In other words, the correct answer must be divisible by 3, which means the sum of its digits must be divisible by 3.
Let's check each answer choice...
A. 2+7+0+4 = 13, which is not divisible by 3. Eliminate.
B. 2+9+9+0 = 20, which is not divisible by 3. Eliminate.
C. 5+4+0+4 = 13, which is not divisible by 3. Eliminate.
D. 5+4+4+4 = 17, which is not divisible by 3. Eliminate.
E. 5+9+9+4 = 27, which IS divisible by 3.
By the process of elimination, the correct answer is E.
General Discussion
Re: There are 27 different three-digit integers that can be formed using o
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25 Apr 2022, 09:00
25 Apr 2022, 09:00
3
Since all the digits whether in the tens, or units place will repeat the same number of times,
i.e. _ _ 1, _ _ 1.... _ _ 2, _ _ 2, ... _ _ 3, _ _ 3 ...
therefore we can say that 9 (1's), 9 (2's), and 9 (3's) will occur since the number of permutations is unbiased to the numbers,
Hence 9x1 + 9x2 + 9x3 = 9 + 18 + 27 = 54,
Now that the sum of the numbers in the unit place is 54, the sum of the numbers in tens will also be 54,
From this we can conclude that the units digit is 4
Hence the carry from the sum of the numbers in the unit place ("5" 4) will be taken to the tens place,
thus the total in the tens digit will be 5+54 = 59, and the 5 gets carried to the next place
therefore eventually we get 5994.
i.e. _ _ 1, _ _ 1.... _ _ 2, _ _ 2, ... _ _ 3, _ _ 3 ...
therefore we can say that 9 (1's), 9 (2's), and 9 (3's) will occur since the number of permutations is unbiased to the numbers,
Hence 9x1 + 9x2 + 9x3 = 9 + 18 + 27 = 54,
Now that the sum of the numbers in the unit place is 54, the sum of the numbers in tens will also be 54,
From this we can conclude that the units digit is 4
Hence the carry from the sum of the numbers in the unit place ("5" 4) will be taken to the tens place,
thus the total in the tens digit will be 5+54 = 59, and the 5 gets carried to the next place
therefore eventually we get 5994.
