Bunuel wrote:
A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?
(A) 18
(B) 19
(C) 21
(D) 42
(E) 59
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:Let’s assume that the 5 natural numbers in increasing order are: a, b, b, b, a+100
We are given that a < b < a+100.
Also, we are given that a and b are positive integers. This information is critical – we will see later why.
The average of the 5 numbers is (a+b+b+b+a+100)/5 = 150
(a+b+b+b+a+100) = 5*150
2a+3b = 650
We need to find the number of distinct values that a can take because a+100 will also take the same number of distinct values.
Now there are two methods to proceed. Let’s discuss both of them.
Method 1: Pure Algebra – Write b in terms of a and plug it in the inequality
b = (650 – 2a)/3
a < (650 – 2a)/3 < a+100
3a < 650 – 2a < 3a + 300
Now split it into two inequalities: 3a < 650 – 2a and 650 – 2a < 3a + 300
Inequality 1: 3a < 650 – 2a
5a < 650
a < 130
Inequality 2: 650 – 2a < 3a + 300
5a > 350
a > 70
So we get that 70 < a < 130. Since a is an integer, can we say that a can take all values from 71 to 129? No. What we are forgetting is that b is also an integer. We know that
b = (650 – 2a)/3
For which values will be get b as an integer? Note that 650 is not divisible by 3. You need to add 1 to it or subtract 2 out of it to make it divisible by 3. So a should be of the form 3x+1.
b = (650 – 2*(3x+1))/3 = (648 – 6x)/3 = 216 – 2x
Here, for any positive integer x, b will be an integer.
From 71 to 129, we have the following numbers which are of the form 3x+1:
73, 76, 79, 82, 85, … 127
This is an Arithmetic Progression. How many terms are there here?
Last term = First term + (n – 1)*Common Difference
127 = 73 + (n – 1)*3
n = 19
a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.
Method 2: Using Transition Points
Note that a < b < a+100
Since a < b, let’s find the point where a = b, i.e. the transition point
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But decreasing a by 3 decreases the largest number i.e. a+100 by 3 too; so we need to increase b by another 1.
We get a = 127 and b = 132. This give us the numbers as 127, 132, 132, 132, 227. Here the average is 150
Since b < a+100, let’s find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But increasing a by 3 increases the largest number, i.e. a+100 by 3 too, so we need to decrease b by another 1.
We get a = 73 and b = 168. This gives us the numbers as 73, 168, 168, 168, 173. Here the average is 150
Values of a will be: 73, 76, 79, ….127 (Difference of 3 to make b an integer)
This is an Arithmetic Progression.
Last term = First term + (n – 1)*Common difference
127 = 73 + (n – 1)*3
n = 19
a will take 19 distinct values so the last term i.e. (a+100) will also take 19 distinct values.
Answer (B)
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