Question

The tenths digit of a certain number is 3 more than the ones digit. The sum of the square of the digit is 29. Find the number.​

Answer 1

Answer:

let the ones digit is x

then tenth digit is (x+3)

so, x^2+(x+3)^2=29

=> x^2+x^2+6x+9=29

=> 2x^2+6x+9-29=0

=> 2x^2+6x-20=0

=> 2(x^2+3x-10)=0

=> x^2+3x-10=0

=> x^2+5x-2x-10=0

=> x(x+5)-2(x+5)=0

=> (x+5)(x-2)=0

=> x+5=0

=> x= -5

x-2=0

=> x=2

x is not negative

so x=2

so ones digit is 2 and tenth digit is (2+3)=5

so the number is 52

Answer 2

Answer:

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Step-by-step explanation:

A two digit number has 3 in its unit digit. The sum of its digits is one seventh of the number itself. What is the number?

Thanks for the A2A!

Assuming base 10, let x be the number in the tens place. Then we can say:

x+3=17(10x+3)

Multiplying both sides by 7 :

7x+21=10x+3

Subtracting 7x from both sides:

21=3x+3

Subtracting 3 from both sides;

3x=18

Dividing both sides by 3 :

x=6

So the number is 10x+3=10(6)+3=63