Math Is Fun Forum

J.

Guest

Help Me!!!!

Can you solve this problem for me with the details of how to solve this:
        The ones digit of a two-digit number is six less than the tens digit. The number is thirty-three more than five times the sum of the digits. What is the number?

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Help Me!!!!

If you call the number 'ab':

Then a = b + 6

as the tens digit is 6 more than the ones digit.

and

10a + b = 33 + 5(a + b)

as the number is 10a plus b (as a is the tens digit and b is the ones digit), and it is equal to the sum of the digits (a+b), multiplied by 5, plus 33.

Rearrange the second equation to give:
10a + b = 33 + 5a + 5b
so 5a - 4b = 33

You know have a pair of simultaneous equations.

So you can substitute the first equation in to the second to give:

5(b+6) - 4b = 33
5b + 30 - 4b = 33
b = 3

If b = 3, and a = b + 6, then a = 9.

So the number is 93.

Last edited by Daniel123 (2007-11-04 07:43:42)

J.

Guest

Re: Help Me!!!!

Can you solve this problem for me with the details of how to solve this:
        The ones digit of a two-digit number is six less than the tens digit. The number is thirty-three more than five times the sum of the digits. What is the number?

Devantè

Real Member

Registered: 2006-07-14

Posts: 6,400

Re: Help Me!!!!

(see Daniel123's post)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help Me!!!!

Yeah, he said 93.
93 = 12 times 5 plus 33.

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