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How many three-digit numbers ABC, in which A, B, and C are each digits 21 Aug 2020, 09:27
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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
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VeritasKarishma chetan2u
What am I doing wrong??

2B = A+C

Range is 111 to 999

B=1
2B=2
A+C= 1+1 or 2+0 (2 combinations)

B=2
2B=4
A+C= 1+3 or 3+1 or 2+2 or 4+0 (4 combinations)

B=3
2B=6
A+C= 1+5 or 5+1 or 2+3 or 3+2 or 3+3 or 6+0(6 combinations)

for B=n where n is values from 1 to 9
there are 2\(n\) number of combination

so total number of combinations are 2+4+6+8.....+16+18

\(\frac{(2+18)}{2}*9\)
= 90
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How many three-digit numbers ABC, in which A, B, and C are each digits 21 Aug 2020, 09:33
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Quote:
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

VeritasKarishma chetan2u
What am I doing wrong??

2B = A+C

Range is 111 to 999

B=1
2B=2
A+C= 1+1 or 2+0 (2 combinations)

B=2
2B=4
A+C= 1+3 or 3+1 or 2+2 or 4+0 (4 combinations)

B=3
2B=6
A+C= 1+5 or 5+1 or 2+3 or 3+2 or 3+3 or 6+0(6 combinations)

for B=n where n is values from 1 to 9
there are 2\(n\) number of combination

so total number of combinations are 2+4+6+8.....+16+18

\(\frac{(2+18)}{2}*9\)
= 90
Show more

There is no pattern in this because as the value of B increases, the number of combinations will start decreasing, as the values of A and B will get restricted
For example :- when B=9, 2B=18 .....A+C =9+9, so just 1, while you calculated 9 for it.
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How many three-digit numbers ABC, in which A, B, and C are each digits 21 Aug 2020, 09:49
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chetan2u

Ah right! The moment 2B becomes a 2 digit number this pattern wont follow and it starts decreasing. Although it decreases with a common difference after B=5---> 2+4+6+8+9+7+5+3+1, I realize solving this way wasn't a good idea.

Thank you for a prompt response. I can sleep peacefully now.
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How many three-digit numbers ABC, in which A, B, and C are each digits 21 Aug 2020, 23:54
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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
Show more


Let us understand why we get 5*5 or 4*5 in our solutions.

1) If you are quick with numbers, then this is the method.

2B=A+C means A+C is even, and for that either both are even or both are odd.
So whenever A+C has a EVEN value, there is a B for it and each such combination will give us a number ABC.
This means we have to concentrate on values of A+C, rather than on B.
(I)
a) Both A and C are ODD..
5 values - 1, 3, 5, 7, 9- possible for each of A and C...Total combinations = 5*5=25
b) Both A and C are EVEN..
5 values - 0, 2, 4, 6, 8 - possible for C, but 0 is not possible for A, so 4 values - 2, 4, 6 and 8....Total combinations = 4*5=20
Answer = 25+20=45

(II)
Total possible values for A without restrictions = 9, that is all digits except 0.
Total possible values for C without restrictions = 10.
All possible combinations = 10*9=90
But possibility of A+C being ODD or EVEN will be same, so 90/2=45 for even.
Answer = 45

2) If we are not very comfortable with permutations and combinations, a normal calculation for values of A, B and C will also help as there are very limited combinations.

B can take values from 0 to 9, but 0 as a value will make ABC as 000.
a) B=1
2B=2......(A,C) = (1,1) and (2,0)..............2
b) B=2
2B=4......(A,C) = (2,2),(3,1),(1,3) and (0,0)..............4
c) B=3
2B=6......(A,C) = (3,3), (4,2)(2,4),(5,1),(1,5) and (0,0).............6
d) B=4
2B=8......(A,C) = (4,4),(3,5),(5,3), (6,2)(2,6),(7,1),(1,7) and (0,0).............8
Now, the sum A+C will become 2-digit number so option (0,0) will not be there
e) B=5
2B=10......(A,C) = (5,5), (6,4),(4,6),(7,3), (3,7),(8,2)(2,8),(9,1),(1,9) .............9
Now each values of B as 6, 7, 8,9 will give 2 lesser than previous, so 7, 5, 3, 1 respectively.
Total \(2+4+6+8+9+7+5+3+1=45\)
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How many three-digit numbers ABC, in which A, B, and C are each digits 22 Aug 2020, 01:01
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2B = A + C

=> B = \(\frac{A + C }{ 2}\)

Now we will get B when the sum of A and C is divisible by 2 or '2' is the factor of the sum of A + C.

Then, A + C has to be even.

Case I: ODD + ODD = EVEN

Case II: EVEN + EVEN = EVEN


Case I: Odd numbers available (1,3,5,7,9) = 5

A = 5 choices and C = 5 choices.

=> Total = 5*5= 25.


Case II: Even numbers available (0,2,4,6,8) = 5

A = 4 choices (starting number will not be '0') and C = 5 choices.

=> Total = 4*5= 20.


Overall ways : 25+20= 45

Answer D
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How many three-digit numbers ABC, in which A, B, and C are each digits 01 Jul 2024, 05:31
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­111
123
135
147 .......999 diffrence 12

so. according to AP formula 999 = 111+(n-1)12 => n =75 . what is the mistake in the logic ? please explain
 
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How many three-digit numbers ABC, in which A, B, and C are each digits 17 Aug 2025, 21:40
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How many three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?

A + C = 2B : A + C is even; Both A & C are odd or both A & C are even; A + C < 10; Since B < 10

Case 1: A & C are both odd
A = 1; C = 1; B = 1
A = 1; C = 3; B = 2
A = 1; C = 5; B = 3
A = 1; C = 7; B = 4
A = 1; C = 9; B = 5
A = 3; C = 1; B = 2
A = 3; C = 3; B = 3
A = 3; C = 5; B = 4
A = 3; C = 7; B = 5
A = 3; C = 9; B = 6
A = 5; C = 1; B = 3
A = 5; C = 3; B = 4
A = 5; C = 5; B = 5
A = 5; C = 7; B = 6
A = 5; C = 9; B = 7
A = 7; C = 1; B = 4
A = 7; C = 3; B = 5
A = 7; C = 5; B = 6
A = 7; C = 7; B = 7
A = 7; C = 9; B = 8
A = 9; C = 1; B = 5
A = 9; C = 3; B = 6
A = 9; C = 5; B = 7
A = 9; C = 7; B = 8
A = 9; C = 9; B = 9
A = {1,3,5,7,9}; C = {1,3,5,7,9}
Total 5*5 = 25 Cases

Case 2: A & C are both even
A = {2,4,6,8}; C = {0,2,4,6,8}
Total 4*5 = 20 cases

The number of three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C = 25 + 20 = 45

IMO D
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