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# 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? $\textbf{(A) } A \qquad \textbf{(B) } B \qquad \textbf{(C) } C \qquad \textbf{(D) } D \qquad \textbf{(E) } E$

#### Number Theory

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 .

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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