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23 Jul 2018, 01:47
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E
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Difficulty:
65%
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Question Stats:
59% (02:21) correct
41%
(02:11)
wrong
based on 758
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GMAT CLUB TESTS' FRESH QUESTION:
{a, b, 1, 2}
If a and b are positive integers less than 10, what is the mode of the list above?
(1) The number of different permutations of the numbers in the list is 12.
(2) A four-digit number 21ab is divisible by 9
M36-111
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23 Jul 2018, 02:07
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Bunuel
Mode means the integer repeated max time
(1) The number of different permutations of the numbers in the list is 12.
All four cannot be different, otherwise permutations would be 4!=4*3*2=24
Since there are 12 permutations, and 12=24/2=4!/2=4!/2!
So there is one pair same and other two distinct..
But mode could be any 1, or 2 or a variable
Insufficient
(2) A four-digit number 21ab is divisible by 9
Since divisibility depends on sum of the number, so a+b+2+1=3+a+b is div by 9
Since a be B are interchangeable, we cannot find the value of variable
Number could 2115,2151,2124,2142,2133,2169,2196,2178,2187
So insufficient
Combined
One pair and other two distinct, so possibilities 2115,2124,2133
So mode can be 1,2 or 3
Insufficient
E
15 Nov 2021, 05:23
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Bunuel
Official Solution:
\(\{a, b, 1, 2\}\)
If \(a\) and \(b\) are positive integers less than 10, what is the mode of the list above?
The mode is the number that occurs the most frequently in a data list. For example, the mode of {2, 3, 4, 4} is 4.
(1) The number of different permutations of the numbers in the list is 12.
If the list contained 4 different numbers, the number of permutations would be \(4!=24\). Since the actual number of permutations is half of that \(\frac{4!}{2}=12\), then the list must contain two identical numbers. The list could be \(\{x, 1, 1, 2\}\), \(\{x, 2, 1, 2\}\) or \(\{x, x, 1, 2\}\) (where \(x\) is different from 1 and 2 ). So, the mode could be 1, 2, or \(a=b=x\). Not sufficient.
(2) A four-digit number \(21ab\) is divisible by 9.
An integer is divisible by 9 if the sum of its digits is divisible by 9. So, \(2+1+a+b=3+a+b\) is divisible by 9. This implies that \(a+b\) must total 6 or 15 (it cannot total more than 15, say 24, because we are told that \(a\) and \(b\) are positive integers less than 10: 1, 2, 3, 4, ..., 9). So, \(a\) and \(b\) can be: (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (6, 9), (9, 6), (7, 8), (8, 7).
If \(a\) and \(b\) are (1, 5) or (5, 1), then the mode will be 1;
If \(a\) and \(b\) are (2, 4) or (4, 2), then the mode will be 2;
If \(a\) and \(b\) are (3, 3), then the mode will be 3;
If \(a\) and \(b\) are other possible values, then there will be no mode (if every number in the list occurs an equal number of times, then the list has no mode. For example list {1, 2, 3} has no mode).
Not sufficient.
(1)+(2) We still cannot determine which number repeats twice in the list. The list could be \(\{1, 5, 1, 2\}\), \(\{2, 4, 1, 2\}\), or \(\{3, 3, 1, 2\}\), giving the mode of 1, 2, or 3 respectively. Not sufficient.
Answer: E
