How many 3-digit integers between 100 and 200 have a digit

Question

How many 3-digit integers between 100 and 200 have a digit that is the average (arithmetic mean) of the other 2 digits?

A. 1
B. 7
C. 9
D. 11
E. 19

BrushMyQuant · Accepted Answer

Let the three digit number be xyz

x can have only one value that is 1 (as its between 100 and 200)
x =  \frac{(y+z)}{2} 
x is 1
=> y + z = 2

=> We will have following cases:
y = 0, z = 2
y = 1, z = 1
y = 2, z = 0

=> We have three numbers as of now
 102, 111, 120

Now considering different values of y
y =  \frac{(x+z)}{2}  and x is 1

So, for y to be integer x+z should be Even
=> z will be odd (as x=1 and Odd + Odd = Even)
=> We can have 5 possible values of z i.e. 1,3,5,7,9
=> Corresponding values of y will be 1,2,3,4,5
=> We have following numbers now
 111, 123, 135, 147, 159

Now considering different values of z
we have z =  \frac{(x+y)}{2}  and x=1
=> z =  \frac{(1+y)}{2}

For z to be an integer we need to have odd values of y
=> y can be 1,3,5,7,9
=> Corresponding values of z will be 
1,2,3,4,5

So we get following numbers
 111, 132, 153, 174, 195

Combining all the cases we will have following numbers:
102, 111, 120, 123,132, 135,147,153,159,174, 195
=> 11

So,  Answer will be D 
Hope it helps!

gurpreetsingh · Answer

IMO d, whats the source? Fix the 1_ _ now for every number 1,2,3,.... (x+y)/2 = 1, = 2.... x+y =2 which is only possible when x=1,y=1 or x=2,y=0 or vice-versa similarly x+y=4 is only possible when x=1 and y=3....or vice-versa, x=3 and y=2 is not possible as one of the x/y has to be 1 :lol:

102,120 111 123,132 135,153 147,174 159,195

nevergiveup · Answer

Since the number is between 100 and 200, the hundreds digit has to be 1. 
And since the answer choices are small enough, it won't hurt just to write them out. 102, 111, 120, 123, 132, 135, 147, 153, 159, 174, 195.

Answer D

beyondgmatscore · Answer

We know that the first digit (hundreth's place) would be 1 for all such numbers. Now second digit (ten's place) need to be an odd number for the third digit to be an integer arithmetic mean, so it can be 1,3,5,7,9 and the third digit would be 1,2,3,4,5 respectively giving us numbers like 111, 132, 153, 174 and 195. By similar logic, we would have numbers where tens and units digit in above numbers interchanged, so 123, 135, 147 and 159.
We also need to worry about cases when the hundred's digit becomes the mean as 1 and in that case, tens and units digit are 0 and 2 and , so we add 102 and 120 to list as well.
So, we have total of 11 numbers. Answer D

panacia · Answer

For jlgdr: 
The first digit is obviously 1. 
A) Let assume that the third digit is the mean of the first and second digits: for the second, we have to choose an odd number to result in an even one when added to 1(first digit), since the digits must be whole numbers(and odd+even results in odd and cannot be divided by 2 to get the other digit). >>> We have 5 options for the second digit. 
B) The other way around is true if assuming the second digit to be the mean. >>> Another 5 options.
NOTE: The 111 is the number that we counted twice in scenario A and scenario B.>>> So far we have 5+5-1=9 options.
C) Lastly, assuming that 1 (first digit) is the mean, we reach 102,120, and 111(counted before). 2 more options
 
>>>> we have 9+2=11 choices. 
Answer is D.
P.S. This approach is not preferable unless you master the process. For me, it took less time than counting. I am not good at counting though! 
Hope it helps...

ScottTargetTestPrep · Answer

Let’s list them: 102, 111, 123, 135, 147, 159 (if the digits, from left to right, are in ascending order). However, we can see that, except for 111, each of the remaining numbers can have their tens and units digits switched. So we have five more numbers: 120,132, 153, 174, 195. Therefore, we have a total of 11 such numbers.

Answer: D

Helium · Answer

This is best done methodically, by considering each of the digits from 0 to 9.

Since each number must be between 100 and 200, there must be a 1 in the hundreds place. If the digit that is the average of the other two digits is also a 1, the other digits must sum to 2. That is, they could be 2 and 0, or 1 and 1. So 111, 102 and 120 are all acceptable.

If the digit that is the average of the other two digits is 2, the other two digits must sum to 4. That is they must be 1 and 3. So 132 and 123 are also acceptable.

If the digit that is the average of the other two digits is 3, then the other two digits must sum to 6, that is they must be 1 and 5. So 135 and 153 are acceptable.

If the digit that is the average of the other two digits is a 4, then the other two digits must be 1 and 7, since they sum to 8. So 147 and 174 are also O.K.

If the digit that is the average of the other two digits is a 5, then the other two digits must sum to 10, that is, they must be 1 and 9. So 159 and 195 are also acceptable.

The digit that is the average cannot be 6 or greater, since the sum of the other two digits would have to be 12 or greater, and you can't get a sum this large if one of the other two digits is 1.

So there are a total of 11 numbers that fit the description in the question stem all together.

Aalto700 · Answer

195 / 159

174 / 147

153

132

120

111

11  hummm

How many 3-digit integers between 100 and 200 have a digit

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How many 3-digit integers between 100 and 200 have a digit

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