Question

a two digit number is equal to 7 times the sum of its digits. the number formed by reversing its digits is less than the original number by 18 . find the original number

Answer 1

The numbers is 46
1st... 4+6= 10
10×7=70
2nd interchange.... 46 recipocal 64
64 - 46 = 18

Answer 2

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$\huge\tt{GIVEN:}$

A two digit number is 7 times the sum of its digits.

The number formed by reversing the digits is 18 less than the original number.

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$\huge\tt{TO~FIND:}$

The number

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$\huge\tt{SOLUTION:}$

Let us assume that the two digits of that number are x and y respectively.

so, The Number formed by them would be, 10x+y

It is also stated that, the number is 7 times the sumSo, we can make an equation by these points as -

$\rule{200}1$

↪(10x + y) ____(EQ.1)

↪10x + y = 7(x+y)

↪10x + y = 7x + 7y

↪10x - 7x = 7y - y

↪3x = 6y

↪x = 6/3 y

↪x = 2y ____(EQ.2)

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The number formed by reversing the digits is, 18 less than the original number

↪10x + y - (10y + x) = 18

↪10x + y - 10y - x = 18

↪9x - 9y = 18

↪9(x-y) = 18

↪(x-y) = 18/9

↪(x-y) = 2 ____(EQ.3)

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Now, putting the value of (EQ.1) to (EQ.2),

↪x - y = 2

↪2y - y = 2

↪y = 2

↪x = 2×2 = 4

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Now, as we got our values,

Putting the values of (EQ.2,3) in (EQ.1)

↪THE NUMBER = (10x + y)

↪THE NUMBER = (10 × 4 + 2)

↪THE NUMBER = (40 + 2)

↪THE NUMBER = 42

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