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

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?

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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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

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

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?

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

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

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

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

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)

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

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!

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 [B= (A+C)/2, a fixed value]

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

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.

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

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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