Largest subset for which the sum of any two numbers is divisible by 46
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Steven G
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What have you tried? Where are you stuck? It is very hard to help you if you do not tell us where you are having trouble.
Here is something to think about. Can you have more than one number less than 46? Can you have any numbers less than 46? Which numbers are divisible by 46?
Have you tried this problem with smaller numbers to learn from them??
Here is something to think about. Can you have more than one number less than 46? Can you have any numbers less than 46? Which numbers are divisible by 46?
Have you tried this problem with smaller numbers to learn from them??
pka
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You should not use the word "Group" when you mean collection. You should have said the \(\displaystyle n\in\mathbb{Z}^+\), a positive integer.
If that is not the case then the given answer is wrong.
I don't know how it is obtained either without thinking about it. And you have said nothing about what ideas you have explored.
Here is one idea to consider. Obviously, the sum of any two numbers divisible by 46 is itself divisible by 46.
[MATH]i,\ j \in \mathbb Z^+ \text { and } i \ne j \implies (i + j) \in \mathbb Z^+ \text { and}[/MATH]
[MATH]46 \text { evenly divides } 46i + 46j \ \because \ \dfrac{46i + 46j}{46} = \dfrac{\cancel {46} (i + j)}{\cancel {46}} = (i + j)..[/MATH]
[MATH]44 * 46 - 2011 = 1840 + 184 - 2011= 2024 - 2011 = 13.[/MATH]
Thus the set of numbers 46 * 1, 46 * 2, ... 46 * 43 is a set of 43 numbers, the sum of any two of which is evenly divisible by 46.
Thus, I suspect one of three things: you have not posted the problem correctly, the answer you have given is wrong, or the set with the maximum number of elements contains numbers that are not multiples of 46.
If the set does contain numbers the sums of which are evenly divisible by 46 but are not themselves divisible by 46, what is the number of elements of the set?[/MATH]
Here is one idea to consider. Obviously, the sum of any two numbers divisible by 46 is itself divisible by 46.
[MATH]i,\ j \in \mathbb Z^+ \text { and } i \ne j \implies (i + j) \in \mathbb Z^+ \text { and}[/MATH]
[MATH]46 \text { evenly divides } 46i + 46j \ \because \ \dfrac{46i + 46j}{46} = \dfrac{\cancel {46} (i + j)}{\cancel {46}} = (i + j)..[/MATH]
[MATH]44 * 46 - 2011 = 1840 + 184 - 2011= 2024 - 2011 = 13.[/MATH]
Thus the set of numbers 46 * 1, 46 * 2, ... 46 * 43 is a set of 43 numbers, the sum of any two of which is evenly divisible by 46.
Thus, I suspect one of three things: you have not posted the problem correctly, the answer you have given is wrong, or the set with the maximum number of elements contains numbers that are not multiples of 46.
If the set does contain numbers the sums of which are evenly divisible by 46 but are not themselves divisible by 46, what is the number of elements of the set?[/MATH]
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Steven G
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Suppose you have two numbers in the form of 46n+a and 46m+b, then a+b =46 (I'm assuming that a and b are positive but less than 46). Now if you have another pair of numbers 46r+c and 46s + d you will have trouble with say 46n+a and 46r+c UNLESS a+c=23.
That is suppose all numbers are an odd multiple of 23
That is suppose all numbers are an odd multiple of 23