word problem type deal, never got how to do these?

Question

A two-digit number is 7 times the sum of its digits. If the digits are reversed, then the new two-digit number is n times the sum of its digits, Find n.

Most thorough explanation gets the 10 =)

Answer 1

Every 2-digit number can be represented as
10t + u, where t is the tens' digit, and u is the units' digit.

First sentence:
10t + u = 7(t+u)

Second sentence:
10u + t = n(t+u)

Simplifying the first sentence: 10t + u = 7t + 7u, so 3t = 6u, so t = 2u

Substitute that into the 2nd sentence:
10u + 2u = n(2u + u)
12u = n(3u)

n = 4

Answer 2

Lets call the original two digit number:
TU
Where T is the tens digit and U is the ones/units digit.
The range of T is the integers 0,1,2,3,4,5,6,7,8,9 (values that the number T could be)
The range of U is the integers 0,1,2,3,4,5,6,7,8,9 (values that the number U could be)

Numerically, this two digit number is really:
10 times T plus U = 10*T + U

The sum of the two digits is T + U. (Sum means addition.)

So, if this two digit number TU is 7 times the sum of its digits T and U:
10*T + U = 7*(T+U)
(sum means addition, times means multiply)

Now, we can expand this out:
10*T + U = 7*(T+U)
10*T + U = 7*T + 7*U
now to group the variables together. We can add/subtract/multiply/divide on both sides of an equation, so long as we do the same thing to both sides, and keep the equality.
10*T + 1*U = 7*T + 7*U
10*T + 1*U - 1*U = 7*T + 7*U - 1*U
10*T = 7*T + 6*U
10*T - 7*T = 7*T + 6*U - 7*T
3*T = 6*U
(3*T)/3 = (6*U)/3
T = 2*U

Now, if we plug in what we know for the value of T = 2*U into the original equation, we can check our work:
10*T + U ?= 7*(T+U)
10*(2*U) + U ?= 7*((2*U) + U)
20*U + U ?= 14*U + 7U
21*U = 21*U


If the digits are reversed to UT, then the this new two-digit number is n times the sum of its digits U and T:
10*U + T = n*(U+T)
subsitute in what we know, T = 2*U
10*U + (2*U) = n*(U + (2*U)
12*U = n*(3*U)
12*U = n*3*U
now, we want to find out the value of n, so we want to get it alone all on one side, so we divide on both sides:
(12*U)/3 = (n*3*U)/3
4*U = n*U
(4*U)/U = (n*U)/U
4 = n

so we never really need to know the values for T and U in order to solve for n, but we could do the trial and error approach... T = 2*U
since we know the range for both T and U are the numbers from 0 through 9, we know that T is an even number (since it is U multiplied by 2), so 0, 2, 4, 6, 8, and that U can be no higher than 4 (since 5*2 is 10, which is larger than the allowed range for T). possible solutions that need to be checked/verified in both of the equations:
pairs (T,U) could be: (0,0), (2,1), (4,2), (6,3), (8,4)
verify by plugging into the original equations:
10*T + U = 7*(T+U)
10*U + T = 4*(U+T)
and all 5 pairs do work in both equations. :)

Answer 3

let x be the digit in the tenth place
let y be the digit in the one place

the sum of the two digits is x + y

the two digit number itself is 10x + y

10x + y = 7(x + y)
10x + y = 7x + 7y
3x = 6y
x = 2y

when the digits are reversed, that is 10y + x
10y + x = n(x + y)

10y + 2y = n (2y + y)
12y = 3yn
n = 4