2015 AMC 8 Problem 23 Number Theory

 Understand the Problem

Tom has twelve slips of paper which he wants to put into five cups labeled $A$, $B$, $C$, $D$, $E$. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$. The numbers on the papers are $2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,$ and $4.5$. If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B$, then the slip with $3.5$ must go into what cup? $\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$
Source of the problem

2015 AMC 8 Problem 23

Topic

Number Theory

Difficulty Level
Easy
Suggested Book
Mathematical Circles

Start with hints

Do you really need a hint? Try it first!

Look at this line ” Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E$.” \( \\ \) And then try to find out the values that will be taken by the cups ultimately .
Notice that sum of the numbers in slips is 35 and the average is 7 . Since the numbers are  consecutive and increasing from $A$ to $E$ . So these are 5 , 6 , 7, 8 , and 9 .
Now its a process of elimination . \( \\ \) Like for the cup A : \( \\ \) We have the sum total for cup A is 5 . Now if the slip with 3.5 goes to cup A then there is no slip for the rest 1.5 . As minimum number assigned for the slips is 2 . So the slip with 3.5 can’t go to A . \( \\ \) Proceed similarly for the rest .
  • Cup B has a sum of 6, but it has told that it already has a 3 slip, leaving \( 6 -3 =3 \), which is smaller for the 3.5  slip. \( \\ \)
Cup C . It must have a value of 7, so adding the 3.5 slip leaves room for \( 7- 3.5 = 3.5 \).But we don’t have the another slip/slips to adjust for 3.5 .[As 2+2 = 4 ] \( \\ \) Cup D . It has a value of 8 . So here so adding the 3.5 slip leaves room for \( 8- 3.5 = 4.5 \) . It can be possilbe as we have the slips with 2 and 2.5  to adjust 4.5 . \( \\ \) But Cup E should also be checked . It has a sum of 9, but it has told that it already has a 2 slip, so we are left with \( 9- 2 = 7 \), which is identical to the Cup C case, and thus also impossible. \( \\ \) So clearly , answer is Cup D .

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