Help with AMC past paper question (1 Viewer)

[Constantspy977]

Im not entire sure how to approach this question, could anyone give working out so I can understand, thank you in advance

 

[CM_Tutor]

It is saying that every integer can be expressed in the form where is zero or a positive integer, is a chosen integer base (we work with all the time), and is the remainder, which is an integer between 0 and . Illustrating with 100:

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You are seeking integers which can be expressed as or as where and are zero or positive integers, where is an integer satisfying , and where .

 

[Constantspy977]

It is saying that every integer can be expressed in the form where is zero or a positive integer, is a chosen integer base (we work with all the time), and is the remainder, which is an integer between 0 and . Illustrating with 100:

​

You are seeking integers which can be expressed as or as where and are zero or positive integers, where is an integer satisfying , and where .

how would you find out how many integers satisfy the equation though? surely its not trial and error

 

[username_2]

how would you find out how many integers satisfy the equation though? surely its not trial and error

From here it is simple enough. As k and m must be integers,
All factors of 15 until 100:
15, 30, 45, 60, 75, 90
All factors of 27:
27, 54, 81

Hence, the number of possible combinations that satisfy the equations are 6*3 = 18 (i think at least)

 

[Constantspy977]

From here it is simple enough. As k and m must be integers,
All factors of 15 until 100:
15, 30, 45, 60, 75, 90
All factors of 27:
27, 54, 81

Hence, the number of possible combinations that satisfy the equations are 6*3 = 18 (i think at least)

but aren't the factors supposed to go a 1000 because that's what it says in the question

 

[username_2]

but aren't the factors supposed to go a 1000 because that's what it says in the question

Oh a 1000 - yep misread the question.

 

[Constantspy977]

For anyone who doesn't know how to do it, I figured it out.
Answer: 30
Why?
Well, you gotta find the lowest common multiple first, which is 15x27, you get 405 and from here to go until 419 because you can only go 14 integers as stated when divided by 15. So you get 405,406,407,408,409,410,411,412,413,414,415,416,417,418,419. Now since you know 405 is a common factor, double it so you know for certain that the doubled number is also a common factor. When you double it, you get 810 and again you only go 14 integers. So 810,811,812,813,814,815,816,817,818,819,820,821,822,823,824. Then you count those numbers and get 30.

Hope this helped

 

[CM_Tutor]

For anyone who doesn't know how to do it, I figured it out.
Answer: 30
Why?
Well, you gotta find the lowest common multiple first, which is 15x27, you get 405 and from here to go until 419 because you can only go 14 integers as stated when divided by 15. So you get 405,406,407,408,409,410,411,412,413,414,415,416,417,418,419. Now since you know 405 is a common factor, double it so you know for certain that the doubled number is also a common factor. When you double it, you get 810 and again you only go 14 integers. So 810,811,812,813,814,815,816,817,818,819,820,821,822,823,824. Then you count those numbers and get 30.

Hope this helped

But, 136 = 9 x 15 + 1 = 5 x 27 + 1.

So, 136 is one of the solutions.

 

[Constantspy977]

But, 136 = 9 x 15 + 1 = 5 x 27 + 1.

So, 136 is one of the solutions.

Oh true, that does make sense, well do you know the best way to figure it out?

 

[CM_Tutor]

You are on the right track, but

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You need to look at cases of integers where is divisible by 5 and is divisible by 9. The smallest possible is then 9, making . You have described above the cases and , but actually can be any integer multiple of 9.

Our solution is that and for each value of , we have 15 possible values of .

The last possible value is when and , that is,

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as the next result would be at .

The solutions come in groups of 15, as you noted, and we now know the starting condition, so up to 1000 we have , by my count... unless I've missed something.