Question

a b c maths puzzle. we have put number from(1to9only) must not use zero in place of alfabetics. the sum must be A B C+ A B C+ A B C=B B B​

Answer 1

Step-by-step explanation:

Given :-

Use the digits except 0 in the places of A,B and C in A B C+ A B C+ A B C=B B B

To find :-

Find A,B and C ?

Solution :-

Given that

A B C+ A B C+ A B C=B B B

A B C

A B C

A B C

_____

B B B

_____

1 4 8

1 4 8

1 4 8

____

4 4 4

____

Explanation :-

Unit digits are C + C + C = B

=> 3 C = B

B is the multiple of 3 .

and

3(A B C ) = B B B

3(100A+10B+C ) = 100B+10B+B

=> 3(100A+10B+C ) = 111B

=>100A + 10B + C = 111B/3

=] 100A+ 10B +C = 37B

=] 100A+C = 37B-10B

=] 100A+C = 27B-----(1)

LHS is more than 100

RHS should be more than 100

By trial and error method

27×1 = 27

27×2 = 64

27×3 = 81

27×4 = 108

So For B = 4 we get 27×4 = 108 which is the first number in the multiple of 27 greater than 100

So , The possible value for B = 4

Then (1) becomes

=> 100A+C = 108

Now we have to find the pairs for A and C which are equal to 108

So, we have

If A = 1 => 100(1)+C = 108

=> 100+C = 108

=> C = 108-100

=> C = 8

If A = 2 the value becomes greater than 108

So,

(A,C) = (1,8) This pair of positive integers is only applicable for 108

So the possible value of A = 1

The possible value of C = 8

So , A = 1, B = 4 ,C= 8

Answer:-

The values of A,B and C are 1,4 and 8 respectively.

Used Method:-

• Trial and Error method