SOLUTION: A two digit number is such that it's value is equal four times the sum of its digits. If the digits are interchanged,the new number formed exceeds two thirds of the original number

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Question 1180014: A two digit number is such that it's value is equal four times the sum of its digits. If the digits are interchanged,the new number formed exceeds two thirds of the original number by 52. Find the original number.
XY =4(a+b)
YX-2/3(4a+b)=52
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52834) About Me  (Show Source):
You can put this solution on YOUR website!
.

I don't know which idea did you try to express or to present writing your equations in the post.

Surely,  it is  (or  "they are")  incorrect and irrelevant. 


The correct equations are


    10a + b = 4(a + b)

    10b + a = %282%2F3%29%2A%2810a+%2B+b%29+%2B+52.


Can you do the rest ?



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


If the tens digit of the number is a and the units digit is b, then the value of the number is 10a+b; and the value of the number with the digits reversed is 10b+a.

Then the two equations are as shown by the other tutor:

10a+b=4(a+b) the value of the number is 4 times the sum of its digits
10b+a=(2/3)(10a+b)+52 the number with its digits interchanged is 52 more than two thirds of the original number

From there, there are still many ways to solve the problem. Below are sketches of a couple of less common ways, involving a little bit of algebra and some easy trial and error.

(1) We can solve the first equation for one variable in terms of the other:

10a+b=4(a+b)
10a+b=4a+4b
6a=3b
2a=b

This tells us the units digit is twice the tens digit. There are not very many 2-digit numbers in which that is true; so check each of them and find which one satisfies the other requirement.

(2) The number is 4 times the sum of its digits, so it is a multiple of 4.

The number with the digits reversed exceeds 2/3 of the original number by 52; that means the original number is a multiple of 3.

So the original number is a multiple of 3*4=12.

Again there aren't many 2-digit multiples of 12; look at them and find the one that satisfies both conditions of the problem.

But probably a formal algebraic solution is required....

So, as the other tutor did, I leave it to you to find a formal algebraic path to the solution.