Finding numbers with substitution/elimination

Question

The difference between a 2 digit number and 5 times the sum of its digits is 18. If the tens digit is twice the units digit, what’s the number?

Answer 1

Let N = number and the tens digit = D₁₀ and the units digit = D_{1 .} Thus N-5(D₁₀+D₁)=18 (given). Next D₁₀=2D₁ (given). Note also that the base 10 number system requires that N=10D₁₀+D₁.

Starting with N-10D₁₀+D₁. substitute 10D₁₀+D₁ for N yielding 10D₁₀+D₁.- 5(D₁₀+D₁) =18= 5D₁₀-4D₁. Substitute 2D₁ for D₁₀ to yield 18=10D₁-4D₁= 6D₁ . Solve to find D₁ =3 and D₁₀=2*3=6. Thus N = 63.

Answer 2

Let t be the tens digit and u be the units digit. Notice that the value of the number is equal to 10t + u. For example, if the number were 24, 10t + u would be equal to 10(2) + 4 = 24.

Now 5 times the sum of the digits can be written as 5(t + u). So the difference between the number and 5 times the sum of its digits is:

10t + u - 5(t + u) = 18.

Simplify:

10t + u - 5t - 5u = 18.

5t - 4u = 18.

Now we also know that the tens digit is twice the units digit, i.e. t = 2u.

Subtract 2u from both sides:

t - 2u = 0.

So now we have a system of two equations:

5t - 4u = 18.

t - 2u = 0.

Let's multiply the second equation by -2 so we can eliminate the u:

5t - 4u = 18.

-2t + 4u = 0.

Add the two equations:

3t = 18.

Divide by 3:

t = 6.

Now we know that t = 2u, so substituting 6 for t we get 6 = 2u. Dividing by two gives us u = 3.

So our number is 63. We can check this easily: five times the sum of the digits is equal to 5*9 = 45. Notice 63 - 45 = 18 just like we wanted. Hope this helps.