Question

cross out the values as you find them.
📩 + 📩 + 📩 + 📩 = 🏠 4 times 📩 must equal a single digit.
📩^📩 = 🐕 now that you know 📩 you can find the dog!
👑 · 🏆 = 🏆 what could the trophy be? anything multiplied by the trophy is the trophy.
🍴 + 🍴 = 🐞 twice the fork is the bug. the bug has to be an even number!
💎 is a prime number. look at the options you have left. which number is prime?

Explanation:

Step1: Find the value of the horse

Let the value of each horse be \( h \). We have \( 5h = 5 \), so \( h=\frac{5}{5}=1 \)? Wait, no, wait. Wait, the first equation: 5 horses add up to 5. So \( h + h+h+h+h = 5 \), so \( 5h = 5 \), so \( h = 1 \)? But looking at the right - hand side, the horse is labeled 8? Wait, maybe I misread. Wait, the right - hand table: horse (blue) = 8? No, wait, let's re - examine. Wait, the first equation: 5 blue horses sum to 5. So each blue horse is \( 5\div5 = 1 \)? But the right - hand table has horse (blue) = 8? Wait, maybe the right - hand table is the key. Wait, the right - hand table: fork = 1, trophy = 2, envelope = 3, hanger = 5, diamond = 4, bug = 5, barn = 6, crown = 7, horse = 8, dog = 9.

Wait, let's start with the crown and trophy equation: crown \(\times\) trophy = trophy. Let crown be \( c \), trophy be \( t \). So \( c\times t=t \). If \( t eq0 \), then \( c = 1 \)? But crown is 7? Wait, no, maybe \( t = 0 \)? But trophy is 2. Wait, maybe the equation is crown \(\cdot\) trophy = trophy, so \( c\times t=t\), which implies \( c = 1 \) (if \( t eq0 \)). But crown is 7. Wait, maybe I made a mistake. Let's look at the fork and bug equation: fork + fork = bug. Fork is 1, so \( 1 + 1=2 \), but bug is 5? No, wait bug is 5. Wait, \( 1+1 = 2 eq5 \). Wait, maybe fork is 2? No, fork is 1. Wait, the diamond is a prime number. Diamond is 4, but 4 is not prime. Wait, diamond is 4, but 4 is not prime. Wait, maybe the diamond is 2? No, diamond is 4. Wait, the envelope equation: 4 envelopes = barn. Envelope is 3, so \( 3\times4 = 12 \), but barn is 6. No, \( 3\times4=12 eq6 \). Wait, envelope is 1? Then \( 1\times4 = 4 eq6 \). Envelope is 2? \( 2\times4 = 8 eq6 \). Envelope is 3? \( 3\times4 = 12 eq6 \). Envelope is 6? No, barn is 6. Wait, \( 4\times1.5=6 \), but envelope should be an integer. Wait, maybe the envelope is 1.5? No, the numbers are 0 - 9 integers.

Wait, let's use the right - hand table. Let's list the right - hand table values:

- Fork (\( f \)) = 1
- Trophy (\( t \)) = 2
- Envelope (\( e \)) = 3
- Hanger = 5
- Diamond (\( d \)) = 4
- Bug (\( b \)) = 5
- Barn (\( r \)) = 6
- Crown (\( c \)) = 7
- Horse (\( h \)) = 8
- Dog (\( g \)) = 9

1. **Crown and Trophy Equation**: \( c\times t=t \). Substitute \( c = 7 \), \( t = 2 \): \( 7\times2=14 eq2 \). This is a problem. Wait, maybe the equation is crown + trophy = trophy? Then \( c=0 \), but crown is 7. No.

2. **Fork and Bug Equation**: \( f + f=b \). \( f = 1 \), so \( 1 + 1=2 eq5 \). \( f = 2 \), \( 2+2 = 4 eq5 \). \( f = 3 \), \( 3 + 3=6 eq5 \). \( f = 5 \), \( 5+5 = 10 \) (not single - digit). Wait, bug is 5, so \( f + f=5 \), then \( f=\frac{5}{2}=2.5 \), not an integer. This is confusing.

Wait, let's go back to the first equation: 5 horses = 5. So each horse is \( 5\div5 = 1 \). But the horse is labeled 8. Maybe the right - hand table is the answer key, and we have to cross out the numbers as we find them.

- From the crown \(\times\) trophy = trophy: If we assume that the only number that when multiplied by another number gives the same number is 1 (for non - zero numbers) or 0 (for zero). But trophy is 2, so crown should be 1, but crown is 7. Alternatively, trophy is 0, but trophy is 2. So maybe the equation is crown \(\cdot\) trophy = trophy implies that crown = 1, but crown is 7. This is a bit confusing.

- From the fork + fork = bug: fork is 1, so bug should be 2, but bug is 5. Fork is 2, bug is 4, but bug is 5. Fork is 3, bug is 6, but bug is 5. Fork is 4, bug is 8, no. Fork is 5, bug is 10, no.

- From the envelope \(\times4=\) barn: envelope is 3, \( 3\times4 = 12 eq6 \). Envelope is 1, \( 1\times4 = 4 eq6 \). Envelope is 2, \( 2\times4 = 8 eq6 \). Envelope is 6, \( 6\times4 = 24 eq6 \).

- From the diamond is a prime number: diamond is 4, which is not prime. Diamond is 2, which is prime, but diamond is 4.

Wait, maybe the numbers to cross out are:

- From 5 horses = 5, each horse is 1, but horse is 8. No.

Wait, the numbers given are 0 - 9, and we have to cross out the values as we find them. Let's list the clues:

1. Crown \(\cdot\) trophy = trophy: The only number \( x \) such that \( x\times y=y \) for \( y eq0 \) is \( x = 1 \). But crown is 7, trophy is 2. So maybe trophy is 0, but trophy is 2. Alternatively, this implies that crown = 1, but crown is 7. So maybe we cross out 1? No, fork is 1.

2. Fork + fork = bug: fork is 1, so bug should be 2, but bug is 5. So \( 1 + 1=2 \), cross out 2? Trophy is 2.

3. Envelope \(\times4=\) barn: envelope is 3, \( 3\times4 = 12 \), no. Envelope is 1, \( 1\times4 = 4 \), diamond is 4. So cross out 4?

4. Diamond is a prime number: diamond is 4, not prime. The prime numbers between 0 - 9 are 2, 3, 5, 7. So if diamond is 4 (not prime), cross out 4?

5. 5 horses = 5: each horse is 1, cross out 1?

This is a bit unclear, but based on the right - hand table and the process of elimination, the numbers to cross out (as we find them) would be:

- From 5 horses = 5: cross out 1 (since each horse is 1)
- From fork + fork = bug: fork is 1, bug would be 2, cross out 2
- From crown \(\cdot\) trophy = trophy: cross out 1 (crown should be 1)
- From envelope \(\times4=\) barn: envelope is 3, \( 3\times4 = 12 \), no. Envelope is 1, \( 1\times4 = 4 \), cross out 4
- From diamond is a prime number: diamond is 4 (not prime), cross out 4

But maybe the intended answer is to cross out the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 one by one as we solve, but the key is to follow the clues.

Wait, let's start over:

1. **Horse**: 5 horses = 5, so each horse is \( 5\div5 = 1 \). But horse is 8? No, maybe the horse is 1, but the table says horse is 8. This is a contradiction. Maybe the first equation is 5 horses = 5, so horse = 1, cross out 1.

2. **Fork and Bug**: Fork + Fork = Bug. Fork is 1, so Bug = 2. Cross out 2.

3. **Crown and Trophy**: Crown \(\times\) Trophy = Trophy. So Crown = 1 (since Trophy \( eq0\)), but Crown is 7. So maybe Trophy = 0, but Trophy is 2. Alternatively, this implies that Crown = 1, cross out 1 (already crossed out).

4. **Envelope and Barn**: 4 Envelopes = Barn. Envelope is 3, 4\times3 = 12 (not single - digit). Envelope is 1, 4\times1 = 4 (Diamond is 4), cross out 4.

5. **Diamond**: Diamond is a prime number. Diamond is 4 (not prime), so the prime numbers left are 2, 3, 5, 7. But Diamond is 4, cross out 4 (already crossed out).

6. **Dog and Envelope**: Envelope to the power of Envelope = Dog. Envelope is 3, \( 3^3=27 \) (not 9). Envelope is 1, \( 1^1 = 1 eq9 \). Envelope is 2, \( 2^2 = 4 eq9 \). Envelope is 4, \( 4^4=256 eq9 \).

This is a bit confusing, but based on the clues and the table, the numbers to cross out (in the 0 - 9 line) as we find them are 1, 2, 3, 4, 5, 6, 7, 8, 9 step - by - step. But the main idea is to use the equations to find the values and cross out the corresponding numbers.

Answer:

The process involves solving each equation to find the values of the symbols and then crossing out the corresponding numbers (1, 2, 3, 4, 5, 6, 7, 8, 9) from the 0 - 9 line as we determine their values based on the given equations (e.g., horse = 1 (cross out 1), fork = 1 (cross out 1), bug = 2 (cross out 2), etc.). The key is to follow the arithmetic relationships between the symbols to identify and cross out the numbers.