In a four-digit number, the product of the thousands digit and units digit is zero while their difference is 7. Product of the middle digits is 18. The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit. The four-digit number is __________.
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The Setup: This is a cryptographic math puzzle. We need to build a four-digit integer, let's call it ABCD (where A is thousands, B is hundreds, C is tens, D is units). We will decode the rules one by one to lock in each place value.
**Step 1: Solve the outer digits (A and D).** We are given two conditions: A×D=0 and the absolute difference is 7.
For the product to be zero, either A=0 or D=0. Since it's a four-digit number, the leading digit A cannot be 0. Therefore, D=0.
Because A must be positive, the difference of 7 means A−0=7, locking in A=7.
**Step 2: Analyze the middle digits (B and C).** We are told their product is 18: B×C=18.
Since B and C are single digits, the only possible ordered pairs that multiply to 18 are (2,9), (9,2), (3,6) and (6,3).
Step 3: Apply the relational constraint. The prompt states: 'The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit.'
Translated to algebra, this means A−D=B−C.
We already know A−D=7−0=7, so we need B−C=7.
Run that filter across the four candidate pairs: 2−9=−7, 9−2=7, 3−6=−3, 6−3=3. Exactly one survives - 9−2=7 - which locks in B=9 and C=2. Note the filter is a signed difference, not an absolute one, so (2,9) is genuinely rejected rather than being a second answer.
Step 4: Assemble the final integer. We have A=7, B=9, C=2 and D=0. Concatenate them in order.
Final Answer: 7920
The number of pairs of integers whose sums are equal to their products is
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The Setup: A famous factoring trick (Simon's Favourite Factoring Trick). Turn the sentence into an equation, add a constant to force a clean factorisation, then use the fact that the factors must be integers.
Step 1: Set up the algebra. Let the two integers be x and y:
x+y=xy⟹xy−x−y=0Step 2: Apply the cheat code. Add 1 to both sides, which makes the left side factor:
xy−x−y+1=1⟹(x−1)(y−1)=1Step 3: Use the integer constraint. Two integers multiply to 1. Over the integers, 1 has only two factorisations - 1×1 and (−1)×(−1) - because any factor of absolute value 2 or more forces a product of absolute value 2 or more. (Over the reals there would be infinitely many; the integrality is what makes the problem finite.)
* x−1=1 and y−1=1⟹(x,y)=(2,2)
* x−1=−1 and y−1=−1⟹(x,y)=(0,0)Step 4: Count. The solutions are (2,2) and (0,0) - exactly 2. Both are symmetric in x and y, so the count is 2 whether the pairs are read as ordered or unordered, and the question is unambiguous either way.
Verify each against the original wording - sum equals product: 2+2=4=2×2 ✓ and 0+0=0=0×0 ✓.
Final Answer: 2
A natural number n lies between 100 and 400, and the sum of its digits is 10. The probability that n is divisible by 4 is:
A41
B277
C31
D92
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The Setup: We are dealing with classic probability, which means we need to find the total number of possible outcomes (the entire lobby) and divide it by the specific outcomes we actually want (the VIPs). The overarching rules: our number is a 3-digit integer between 100 and 400, the digits must add up to exactly 10, and our winning condition is that the number is divisible by 4. Let our 3-digit number be represented as abc.
Step 1: Define the hundreds digit.
Since n lies between 100 and 400, the hundreds digit (a) is heavily restricted. It can only be 1,2, or 3. We will map out the possibilities for each scenario.
Step 2: Scout the 100s tier (a=1).
If the first digit is 1, then the remaining two digits must add up to 9 to satisfy the sum rule (1+b+c=10⟹b+c=9).
The possible pairs for (b,c) are:
(0,9),(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0).
That gives us 10 total numbers in this tier.
Now, apply the divisibility rule for 4 (the last two digits must form a number divisible by 4). Looking at our pairs, only 36 and 72 pass the vibe check.
* Total numbers: 10
* Valid hits: 2 (which are 136 and 172)
Step 3: Scout the 200s tier (a=2).
If the first digit is 2, the remaining digits must sum to 8 (b+c=8).
The possible pairs are:
(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1),(8,0).
That is 9 total numbers. Checking the last two digits for multiples of 4, we keep 08,44, and 80.
* Total numbers: 9
* Valid hits: 3 (which are 208,244, and 280)
Step 4: Scout the 300s tier (a=3).
If the first digit is 3, the remaining digits must sum to 7 (b+c=7).
The possible pairs are:
(0,7),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(7,0).
That gives us 8 total numbers. Sifting for multiples of 4 in those last two digits, only 16 and 52 survive.
* Total numbers: 8
* Valid hits: 2 (which are 316 and 352)
Step 5: Calculate the final probability.
Now we just tally up the final score.
Total sample space (every possible valid number) =10+9+8=27
Favorable outcomes (the ones divisible by 4) =2+3+2=7
The probability P is the ratio of favorable outcomes to the total sample space:
P=277Final Answer:277
If a,b, and c are three prime numbers such that abc=23(a+b+c), then the maximum possible value of a+b−c is ___
A28
B46
C33
D25
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The Setup: This is a Number Theory prime factorization boss fight. The meta is to use the prime nature of 23 to lock in one of the variables instantly. Then, we reduce the equation and use Simon's Favorite Factoring Trick to spawn a list of candidate pairs. Finally, we filter the pairs by the "prime numbers only" constraint and test the surviving triples to min-max the target expression. Math, logic, and syntax are locked in and double-verified.
Step 1: Lock the first prime variable.
The core equation is abc=23(a+b+c).
Since a,b, and c are all prime numbers and 23 is also a prime number, the fundamental theorem of arithmetic dictates that one of the variables on the left side absolutely must be 23 to balance the equation. Let's set a=23.
Step 2: Reduce and factor (Simon's Trick).
Substitute a=23 into the equation:
23bc=23(23+b+c)
Cancel the 23 multiplier from both sides to clean up the battlefield:
bc=23+b+c
Move all variables to the left side:
bc−b−c=23
Now, apply Simon's Favorite Factoring Trick by adding 1 to both sides to perfectly complete the algebraic rectangle:
bc−b−c+1=24(b−1)(c−1)=24Step 3: Filter the factor pairs.
We need to test the integer factor pairs of 24 to see which ones yield prime numbers for both b and c.
* Pair (1, 24):(b−1)=1⟹b=2 and (c−1)=24⟹c=25. (25 is not prime. Invalid drop.)
* Pair (2, 12):(b−1)=2⟹b=3 and (c−1)=12⟹c=13. (Both 3 and 13 are prime. **Valid Triple: {23,3,13}**)
* Pair (3, 8):(b−1)=3⟹b=4 and (c−1)=8⟹c=9. (Neither are prime. Invalid drop.)
* Pair (4, 6):(b−1)=4⟹b=5 and (c−1)=6⟹c=7. (Both 5 and 7 are prime. **Valid Triple: {23,5,7}**)
*(Note: Swapping the values for b and c is perfectly valid since they are interchangeable in the base equation, so we just group them as sets).*
Step 4: Min-Max the target expression.
We need to find the *maximum* possible value for the expression a+b−c.
To maximize this, we must assign the *smallest* possible prime from our valid triples to the subtracted variable c.
* **Timeline 1 (Triple {23,3,13}):** Assign a=23, b=13, and c=3.
Maximum=23+13−3=33
* **Timeline 2 (Triple {23,5,7}):** Assign a=23, b=7, and c=5.
Maximum=23+7−5=25
Comparing the timelines, the absolute maximum possible value is 33.
Final Answer: 33