Question

Topic-Playing with Numbers
(See the attachment for question)


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Answer 1

$\large\text{\underline{Let's begin:-}}$

This is a maths puzzle where an alphabet corresponds to a single-digit number. To find it, we use carry-over or digit guessing, etc.

$\large\text{\underline{Solution:-}}$

First, let's compare the number in subtraction.

In the first subtraction,

$\hookrightarrow3\ \underline{C}\ -2\ 7\ =\ 6$

So, $C=3$.

In the second subtraction,

$\hookrightarrow 6\ \underline{D}\ -\ 5\ 4\ =8$

We can find that $D=2$.

Then, let's compare the number in the product.

$\hookrightarrow \underline{A}\ \underline{B}\ \times\ 1\ =2\ 7$

Hence, we find that $A=2$ and $B=7$.

Now integrating the steps,

$\hookrightarrow 3\ 3\ 2\ \underline{E}\ \div\ 2\ 7\ =\ 1\ \underline{F}\ \underline{G}$

Observing the equation, there is no remainder. Hence, $3\ 3\ 2\ \underline{E}$ is divisible by 27, hence shows 9 must divide it.

By divisibility rule,

$\hookrightarrow 3+3+2+\underline{E}=9$

$\hookrightarrow \underline{E}+8=9$

$\hookrightarrow \underline{E}=1$

Now it is time to use the division equation.

$\hookrightarrow 3\ 3\ 2\ 1\ \div\ 2\ 7\ =\ 1\ \underline{F}\ \underline{G}$

$\hookrightarrow 1\ \underline{F}\ \underline{G}\ \times\ 2\ 7\ =\ 3\ 3\ 2\ 1$

In the unit digit, we see that the product of 7 and an unknown number ends with 1. The only possibility is that,

$\hookrightarrow \underline{G}=3$

Now,

$\hookrightarrow 1\ \underline{F}\ 3\ \times\ 2\ 7\ =\ 3\ 3\ 2\ 1$

$\hookrightarrow \underline{F}=2$

$\large\text{\underline{Conclusion:-}}$

$A=2,B=7,C=3,D=2,E=1,F=2,G=3$ is the answer.

Answer 2

Given :-

AB) 3 C D E (1FG

    - 27

______________

        6 D

      -  5 4

_________________

                     8 E

                    - 8 E

___________________

                            x

To Find :-

Value of A, B, C, D, E, F and G

Solution :-

For C

3 C - 27 = 6

3 C = 27 + 6

3 C = 33

Taking the ones digit

C = 3

C = 3

For D

6 D - 54 = 8

6 D = 54 + 8

6 D = 62

Taking the ones digit

D = 2

Now

For A and B

AB × 1 = 27

AB = 27/1

AB = 27

Tens digit will be A and ones digit will be B

A = 2 & B = 7

For E

332E/27 = 1FG

332E/27 = FG

The Divisibility rule of 9 states that the number is divisible by 9. If the sum of all digits of the number is 9 or a multiple of 9

3 + 3 + 2 + E = 9

8 + E = 9

E = 9 - 8

E = 1

From F and G

FG × 27 = 3321

27FG = 3321

Take the ones and tens digit

FG = 21

F = 2 & G = 1