Question

A number consists of two digits of which tens digit exceeds the unit digit by $7$. The number itself is equal to $10$ times the sum of its digits. Find the number.

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Answer 1

Let the two digit number be : AB

Given : Ten's digit exceeds the unit digit by 7

$\longrightarrow$  A = B + 7

Given : The Number itself is equal to 10 times the sum of its digits

$\longrightarrow$  AB = 10(A + B)

★  AB can be written as : 10A + B

(Because : A is in Ten's place and B is in One's place)

$\longrightarrow$  10A + B = 10A + 10B

$\longrightarrow$  9B = 0

$\longrightarrow$  B = 0

$\longrightarrow$  A = (0 + 7) = 7

Answer : The Two Digit Number is 70

Answer 2

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Let the number be xy, where x is the digit at ten's place and y is the digit at one's place.

Now, it is given that

==> x=y+7. -----------> (1)

Also, we know that

A two digit number can be written as

==> 10A+B, where A is ten's digit and B is one's digit

So, our number is 10x+y.

According to question,

==> 10x+y=10(x+y)

==> 10x+y=10x+10y

==> y=0

Now, from (1)

x=y+7

x=0+7

x=7

Hence, the number is 10(7)+0=70.

This is the best possible answer

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