Solving the Sisters' Toy Distribution Problem
The problem involves tracking the distribution of 24 toys among three sisters (R, S, T) through several exchanges, ending with equal shares. The key to solving this is to work backward from the final state.
Final Distribution Analysis
Total toys = 24. At the end, the three sisters have an equal number of toys. Therefore, each sister has $\frac{24}{3} = 8$ toys.
Working Backward: Step-by-Step Reconstruction
We reverse the transactions to find the original distribution.
Step 3: Reversing T's Gift to R
Final State: R=8, S=8, T=8.
This step involved T giving toys to R, which doubled R's share. This means just before this exchange, R had half the toys she had at the end.
- R's toys before T gave: $8 \div 2 = 4$ toys.
- The number of toys T gave to R is $8 - 4 = 4$ toys.
- T's toys before giving them to R were her final share plus what she gave away: $8 + 4 = 12$ toys.
- S's share remained unchanged during this specific transaction: $8$ toys.
State before Step 3: R = 4, S = 8, T = 12.
Step 2: Reversing S's Gift to T
This step involved S giving toys to T, which doubled T's share. So, before this exchange, T had half the toys she had at the end of Step 2 (which is the state before Step 3).
- T's toys before S gave: $12 \div 2 = 6$ toys.
- The number of toys S gave to T is $12 - 6 = 6$ toys.
- S's toys before giving them to T were her share at the end of Step 2 plus what she gave away: $8 + 6 = 14$ toys.
- R's share remained unchanged during this specific transaction: $4$ toys.
State before Step 2: R = 4, S = 14, T = 6.
Step 1: Reversing R's Gift to S
This step involved R giving toys to S, which doubled S's share. So, before this exchange, S had half the toys she had at the end of Step 1 (which is the state before Step 2).
- S's toys before R gave: $14 \div 2 = 7$ toys.
- The number of toys R gave to S is $14 - 7 = 7$ toys.
- R's toys originally were her share at the end of Step 1 plus what she gave away: $4 + 7 = 11$ toys.
- T's share remained unchanged during this specific transaction: $6$ toys.
Original State: R = 11, S = 7, T = 6.
Original Number of Toys for R
Based on the backward calculation, R originally had 11 toys.
Verification
Let's verify the forward process with the original state (R: 11, S: 7, T: 6):
- R gives 7 toys to S (doubling S's share). State becomes R: (11-7)=4, S: (7+7)=14, T: 6. Total = 4+14+6=24.
- S gives 6 toys to T (doubling T's share). State becomes R: 4, S: (14-6)=8, T: (6+6)=12. Total = 4+8+12=24.
- T gives 4 toys to R (doubling R's share). State becomes R: (4+4)=8, S: 8, T: (12-4)=8. Total = 8+8+8=24.
The final state is equal shares (8 toys each), confirming the original calculation.
Important Questions from Numerical Reasoning
- $P, Q, R, S, X$, and $Y$ are distinct single-digit whole numbers taking values from 0 to 9.
$PQ$ is a two-digit number with $Q$ being in the units place and $P$ in the tens place. Similarly, $RS$ is a two-digit number.
It is known that $PQ$ and $RS$ are consecutive numbers and
$(PQ)^2 + (RS)^2 = XYP$, with $XYP$ being a three-digit number.
The value of $Y$ is __________ - Let $p_1$ and $p_2$ denote two arbitrary prime numbers. Which one of the following statements is correct for all values of $p_1$ and $p_2$?
If $\oplus \div \odot = 2$, $\oplus \div \triangle = 3$, $\odot + \triangle = 5$, and $\Delta \times \otimes = 10$,
then the value of $(\otimes - \oplus)^2$ is:- The remainder when $98!$ is divided by $101$ is equal to ________
- A 'frabjous' number is defined as a 3 digit number with all digits odd, and no two adjacent digits being the same. For example, 137 is a frabjous number, while 133 is not. How many such frabjous numbers exist?
