How many three-digit numbers ABC, in which A, B, and C are each digits

Question

How many three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?

A. 33
B. 36
C. 41
D. 45
E. 50

vitaliyGMAT · Accepted Answer

2B=A+C is the same as B=(A+C)/2. That is B is the arithmetic mean of the other 2 digits. 
This can be achieved only in case, when parity of both A and C is the same. So we have odd + odd or even + even.
Let's look at first case. We have 5 ways to choose odd number for A and same 5 ways to choose odd number for C - total 5*5=25.
In the case of even numbers, we can't choose 0 for A (otherwise it won't be 3 digit number) so we have 4*5=20. Those cases are independent so in total we have 45 ways.
Answer is D.

KarishmaB · Answer

Alternatively, we are given that 
2B = A + C
So whatever A and C are, they add up to give an even sum. Hence either A and C are both even or both odd. 
Now proceed as given above.

paidlukkha · Answer

first 3 digit number which qualifies for 2B=A+C > 111
next is 123
next is 135
gap of 12 
900 3 digit numbers 
so 900/12=50

E?

ramakant · Answer

Possible solutions are
Diff of 0 -- 
111,222,........,999 = 9
Diff of 1 --
123, 234, ....., 789 =7x2 = 14
Diff of 2 --
135, 246.......579 = 5x2 = 10
Diff of 3 --
147,258,369 = 3x2 = 6
Diff of 4 --
159 = 1x2 = 2
Last
210, 420,630,840 = 4

So total is 9+14+10+6+2+4 = 45.

Not a good way to solve.
Experts please

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mbaprep2016 · Answer

2B = A+C 
Middle number can only be 2, 4,6,8,10,12,14,16,18
now when middle number is 2 , you can have only 1 combination A =1 and C =1

when middle is 4 you can have two pairs 1 ,3 and 2,2
when middle is 6 you can have three 
so basically 1+2+3+4+5+6+7+8+9 = 45
D is the answer

I ma confused about one thing , why we are not required to consider op pair
when we have 4 at middle we can have (1,3 ) (2,2)and (3,1)
when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51 
another series is 1+3+5+...17 == 81 , which is not answer

prady80 · Answer

2B = A + C
Let A=C
therefore 2A = 2B 
therefore A=B=C is part of solution.
There are 9 such occurrences

now let A = 1, possible values of c =3,5,7,9
 A = 2, possible values of c =4,6,7,0
 A = 3, possible values of c =1,5,7,9
 A = 4, possible values of c =2,6,8,0
... so on...
There are 9 such rows = 9 * 4 = 36
Total = 36 + 9 = 45
D

MorningRunner · Answer

I just did manual combinations.. Though I got it right, it took lot of time..
A-B-C
Start with number 9 for B..

A-9-C --> we have to find combinations of A and C such that A+C=18 ----> this yields only  one  combination --> 9-9-9
A-8-C --> we have to find combinations of A and C such that A+C=16 ----> this yields  three  combinations --> 8-8-8, 9-8-7, 7-9-9
..
..
..
..

Do the same for all digits and make the list
B----  num of combinations 
9----  1 
8----   3 
7----   5 
6----   7 
5----   9 
4----   8 
3----   6 
2----   4 
1----  2 
0----  0 
-------------
Total ----  45 (add all above numbers in red color)

Answer choice D is correct..
But this procedure takes lot of time...

Is there any quick solution?

gracie · Answer

If we assume numbers that work are distributed 
equally through the 9 100s blocks,
only answers B and D are multiples of 9.
If we choose any 100s block, say the 500s, and start at 555,
which we know will work, and count forward and backward
by intervals of 12 until numbers no longer satisfy,
we will have 5 numbers that work: 555,567,579,543,531.
5*9=45
D

SVaidyaraman · Answer

1.Interpret what is given so that we get basic info. 2B = A + C implies that A and C are both odd or both even
2. When A and C are both odd, A can be 1,3,5,7,9 and for each of these B can be 1,3,5,7,9 for a total of 25 cases
3. When A and C are both even A can be 2,4,6,8 and for each of these B can be 0,2,4,6,8 for a total of 20 cases
4. So the answer is 25+20= 45 cases

GMATisLovE · Answer

I also have the same doubt. Can some expert help here?

SVaidyaraman · Answer

Hi,

We have 2B=A+C as follows:

when middle number is 1 we have the pairs (1,1) (2,0)
When middle number is 2 we have the pairs (2,2 ) (3,1) (1,3 ) (4,0)
When middle number is 3 we have the pairs (3,3) (4,2), (2,4) (5,1) (1,5) (6,0)
When middle number is 4 we have the pairs (4,4 ) (5,3) (3,5) (6,2) (2,6) (1,7) (7,1) (8,0)
When middle number is 5 we have the pairs (5,5) (6,4,) (4,6) (7,3) (3,7) (8,2) (2,8) (9,1) (1,9)
When middle number is 6 we have the pairs (6,6 ) (7,5) (5,7) (4,8 ) (8,4 ) (9,3) (3,9 ) 
When middle number is 7 we have the pairs (7,7) (8,6) (6,8) (9,5) (5,9)
When middle number is 8 we have the pairs (8,8 ) (9,7) (7,9)
When middle number is 9 we have the pairs ( 9,9)

for a total of 45 cases

hellosanthosh2k2 · Answer

Given : ABC is three digit number, so A > 0
 A + C = 2B => A + C = even

case 1: both A and C are odd
 odd digits: {1,3,5,7,9} = there are 5 * 5 possible permutations => total = 25

case 2: both A and C are even
 even digits : {0,2,4,6,8} => possible digits for A : {2,4,6,8} = 4
 possible digiits for C: {0,2,4,6,8} = 5
 total permutations : 4 * 5 = 20

so number of such three digit number = 25 + 20 = 45 => (D)

Aprajita760 · Answer

Hi,
How are you getting 4*5?
Please explain.

GMATMBA5 · Answer

Given that 2B=A+C, that is, B =(A+C)/2

For A+C to be divisible by 2, A+C should be even. And it will be even only when both A and C are either odd or even at the same time. Recall even odd rules.(E+O=O)

Odd values A and C can hold - 1, 3, 5, 7 and 9 - that is 5 values each. Applying basic selection property we get total possible selections for odd A and C as - 5*5= 25 numbers

Even values A can hold - 2, 4, 6, 8 = 4 values
Even values C can hold - 0, 2, 4, 6, 8 = 5 values
Total selections in case for even A and C = 4*5=20 numbers

Therefore total numbers both even and odd = 25+20 = 45 numbers

---------------------------------------------------
 Kudos if helpful!

Shobhit7 · Answer

Digit A has 9 options (1-9)
All digits except 0

Digit C has 5 options (either odd or even)
Reason: If A is odd, C is odd & If A is even, C is even; That's because if A and C are opposites, then their sum will be odd and (A+C)/2 = B will become a non integer.

Digit B has 1 option

Total possible three digit numbers= 9x5x1 = 45
Ans D

EncounterGMAT · Answer

SravnaTestPrep 
Hello sir, I have a doubt in the highlighted part. Since zero is even, why zero is not considered as one of the value of A? As in if at all I divide 2 on both the sides of the equation given in the question stem, it would be B=A+C/2 => so both A & C can either be even or odd. I did the calculation this way and ended up getting 50 as answer. Pls guide.

Shobhit7 · Answer

EncounterGMAT

A represents Hundred digit place in three digit number ABC.

If A is taken as 0, then the number will change from a three digit to a two digit number.

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EncounterGMAT · Answer

Thank you. My mind drifted so much into the calculations that I forgot about the digits.

kajaldaryani46 · Answer

Okay so my method of calculating was different from the above - I got the answer but I’m not sure it’s entirely correct:

Since the 3 digit no is ABC, I rewrote it as 100A+10B+C
Now 2B=A+C
So 2(10)B = 100 A + C
Dividing both sides by 20, I get:
B=5A+ C/20
So if all are integers, C has to be a multiple of 20
Between any 100 numbers (eg 1-100), you have 5 nos divisible by 20
So 100- 1 case divisible by 20
101-200 - 5 cases
201-900- 5*7 cases
901- 999 - 4 cases
Total = 45 cases
But I haven’t done anything for A because that could take any value - so I’m not sure if this was the right method. Can somebody please advise?

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How many three-digit numbers ABC, in which A, B, and C are each digits

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