Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 2 | Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
please help!
Date Posted: 3 years ago
Views: 435
See 1 Answer
225m - 169n = 24, m,n > 0
One thing is for sure : m < n
[Because, if m = n, then 225m - 169n = 225m - 169m = 56m. And since m > 0, then the difference will be a multiple of 56, which is bigger than 24 already.
If m > n, the difference will even be bigger than that multiple of 56 previously.]
Next, let n = m + a, where a is a positive integer.
225m - 169(m + a) = 24
56m - 169a = 3 x 8
8(7m) - 3 x 8 = 169a = (13 x 13)a
8(7m - 3) = (13²)a
2³(7m - 3) = 13²a
Now notice both sides have only prime factors now (other than m and a, which are unknown still.)
Since prime numbers have no factors other than 1 and itself, then the 13² and 2³ have no other factors in them.
Both sides are products of factors that give the same number. In order for both sides to be equal, that 7m - 3 has to be a multiple of 13² (=169). We can't factor 2 out from 13, so a must be a multiple of 2³ (= 8)
So we should first find the multiple of 169. Now 169 ÷ 7 = 24 R 1
This means that in each 169, we can have 24 sets of 7, and a 1. If we have multiples of 169, each of their remainders of 1 can be grouped together. As long as adding 3 to their sum gives us a multiple of 7, we would have satisfied 7m.
The smallest number we can add 3 to to get a multiple of 7, is 4. (4 + 3 = 7)So we need a minimum of four 169s.
4 x 169 = 7m - 3
676 = 7m - 3
7m = 679
m = 97
Then, a = 2³(7m - 3)/13² = 8(676)/13² = 32
n = 97 + 32 = 129
∴ a possible solution is m = 97, n = 129
One thing is for sure : m < n
[Because, if m = n, then 225m - 169n = 225m - 169m = 56m. And since m > 0, then the difference will be a multiple of 56, which is bigger than 24 already.
If m > n, the difference will even be bigger than that multiple of 56 previously.]
Next, let n = m + a, where a is a positive integer.
225m - 169(m + a) = 24
56m - 169a = 3 x 8
8(7m) - 3 x 8 = 169a = (13 x 13)a
8(7m - 3) = (13²)a
2³(7m - 3) = 13²a
Now notice both sides have only prime factors now (other than m and a, which are unknown still.)
Since prime numbers have no factors other than 1 and itself, then the 13² and 2³ have no other factors in them.
Both sides are products of factors that give the same number. In order for both sides to be equal, that 7m - 3 has to be a multiple of 13² (=169). We can't factor 2 out from 13, so a must be a multiple of 2³ (= 8)
So we should first find the multiple of 169. Now 169 ÷ 7 = 24 R 1
This means that in each 169, we can have 24 sets of 7, and a 1. If we have multiples of 169, each of their remainders of 1 can be grouped together. As long as adding 3 to their sum gives us a multiple of 7, we would have satisfied 7m.
The smallest number we can add 3 to to get a multiple of 7, is 4. (4 + 3 = 7)So we need a minimum of four 169s.
4 x 169 = 7m - 3
676 = 7m - 3
7m = 679
m = 97
Then, a = 2³(7m - 3)/13² = 8(676)/13² = 32
n = 97 + 32 = 129
∴ a possible solution is m = 97, n = 129
Some notes :
For 7m - 3, m can actually be other values.
Since 7m - 3 divides 169 and we know our minimum m is 97, as long as we increase our m by multiples of 169, the end result of 7m - 3 will still be divisible by 169.
Eg. If m = 97 + 169k, k is a integer and k > 0.
7m - 3 = 7(97 + 169k) - 3
= 7(97 - 3) + 7(169k)
=169 x 4 + 169(7k)
= 169(4 + 7k)
Then, a = 169(4 + 7k)(2³) / 13²
= 8(4 + 7k)
= 32 + 56k
n = 97 + 169k + 32 + 56k
= 129 + 225k
Also realise that the initial equation
225m - 169n = 24 is in the form of the equation of a straight line (y = mx + c)
i.e 225m = 169n + 24 → m = 169/225 n + 24 / 225
So there are multiple values that satisfy the equation.
Sub m = 169k + 97 and n = 225k + 129 into the original equation,
225(169k + 97) - 169(225k + 129) = 24
225(169k) - 169(225k) + 225(97) - 169(129) = 24
Notice that the 225(169k) - 169(225k) cancels out.
This works as 225 = 5² x 3² and 169 = 13², the LCM is 5² x 3² x 13² = 225 x 169
no matter which k you choose, it is always going to cancel each other out.
For 7m - 3, m can actually be other values.
Since 7m - 3 divides 169 and we know our minimum m is 97, as long as we increase our m by multiples of 169, the end result of 7m - 3 will still be divisible by 169.
Eg. If m = 97 + 169k, k is a integer and k > 0.
7m - 3 = 7(97 + 169k) - 3
= 7(97 - 3) + 7(169k)
=169 x 4 + 169(7k)
= 169(4 + 7k)
Then, a = 169(4 + 7k)(2³) / 13²
= 8(4 + 7k)
= 32 + 56k
n = 97 + 169k + 32 + 56k
= 129 + 225k
Also realise that the initial equation
225m - 169n = 24 is in the form of the equation of a straight line (y = mx + c)
i.e 225m = 169n + 24 → m = 169/225 n + 24 / 225
So there are multiple values that satisfy the equation.
Sub m = 169k + 97 and n = 225k + 129 into the original equation,
225(169k + 97) - 169(225k + 129) = 24
225(169k) - 169(225k) + 225(97) - 169(129) = 24
Notice that the 225(169k) - 169(225k) cancels out.
This works as 225 = 5² x 3² and 169 = 13², the LCM is 5² x 3² x 13² = 225 x 169
no matter which k you choose, it is always going to cancel each other out.
360 Tutoring Program