Answer: The loser's card shows 6.
Explanation: Let's start by naming the first student A and the second student B.
Since the product of A and B are either 12, 15, or 18, let's list every single possibility, the first number being A's number and the second number being B's number.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
4 3
5 3
6 2
6 3
9 2
12 1
15 1
18 1
Now, the information says that A doesn't know what B has, so we can immediately cross off all of the combinations that have the integer appearing once and once ONLY off, because if it happened once only, A would know of it straight away. Now, our sample space becomes much smaller.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
6 2
6 3
Using this same logic, we know that we can cross off all of the digits that occur only once in B's column.
2 6
3 6
Now, A definitely knows what number B has because there is only one number left in B. Hence, we can conclude that the loser, B, has the integer 6.
You are not dumb. No one is dumb. Einstein failed math in high school.
Anna is x years old.
Jane is y years younger.
This means the Jane is x - y years old.
In three years, Anna will be x + 3 years old.
In three years, Jane will be (x - y) + 3 years old.
Let S = the sum of their ages in 3 years.
S = x + 3 + (x - y) + 3
S = 2x + 6 - y
Got it?
4d + 9 - d - 8 = (4d - d) + (9 - 8) = d(4 - 1) + 1 = 3d + 1
Answer:
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Step-by-step explanation:
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