CAT 2025 Slot 2QANumber Systems • Integral SolutionsHard
Suppose a,b,c are three distinct natural numbers, such that 3ac=8(a+b). Then, the smallest possible value of 3a+2b+c is
TITA Answer:
⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution
Official Correct Answer: 12. First, We analyze the equation 3ac = 8(a + b). Maybe I can rearrange it to express one variable in terms of the others. Let me try solving for b. If I do that, I can subtract 8a from both sides and then divide by 8. So, 3ac - 8a = 8b, which simplifies to b = (3ac - 8a)/8. Hmm, that might not be the most helpful form. Maybe Factoring out an a from the numerator: b = a(3c - 8)/8. Since b has to be a natural number, the numerator must be divisible by 8. So, a(3c - 8) must be a multiple of 8.
Let me think about this. Since a and c are natural numbers, 3c - 8 must be positive because b has to be positive. So, 3c - 8 > 0, which means c > 8/3. Since c is a natural number, the smallest c can be is 3. Wait, 8/3 is approximately 2.666, so c must be at least 3.
Now, let's consider possible values of c starting from 3 and see what a and b would be. Maybe I can express a in terms of c or vice versa. Let's rearrange the original equation: 3ac = 8a + 8b. If I factor out a, I get 3c = 8 + 8b/a. Hmm, that might not be helpful. Maybe I can express a in terms of c and b. 3ac = 8a + 8b => 3ac - 8a = 8b => a(3c - 8) = 8b. So, a = (8b)/(3c - 8). Since a must be a natural number, (8b) must be divisible by (3c - 8).
This seems a bit complicated. Maybe I can approach it differently. Let's consider that 3ac = 8(a + b). Let's solve for c: c = (8(a + b))/(3a). Since c must be a natural number, 8(a + b) must be divisible by 3a. So, 8(a + b) is a multiple of 3a. Let's write this as c = (8/3)(1 + b/a). Hmm, since c is an integer, (1 + b/a) must be a multiple of 3/8. Wait, that might not be straightforward.
Alternatively, let's consider that 3ac = 8(a + b). Let's write this as 3ac - 8a = 8b => a(3c - 8) = 8b. So, a must be a divisor of 8b. But since a and b are distinct, we need to find a such that a divides 8b, but a ≠ b. Hmm, this is getting a bit tangled.
Maybe I can express b in terms of a and c. From the equation, b = (3ac - 8a)/8. Let's factor out a: b = a(3c - 8)/8. Since b must be a natural number, (3c - 8) must be such that when multiplied by a, it's divisible by 8. So, (3c - 8) must be a multiple of 8/a. Since a is a natural number, 8/a must be an integer, meaning a must be a divisor of 8. So, possible values of a are 1, 2, 4, 8.
Let's test these possible a values one by one.
Starting with a = 1:
Then, b = (3c - 8)/8. Since b must be a natural number, (3c - 8) must be divisible by 8. So, 3c ≡ 8 mod 8 => 3c ≡ 0 mod 8. So, 3c must be a multiple of 8. Since 3 and 8 are coprime, c must be a multiple of 8. Let's let c = 8k, where k is a natural number. Then, b = (3*8k - 8)/8 = (24k - 8)/8 = 3k - 1. Since b must be a natural number, 3k - 1 ≥ 1 => k ≥ 1. So, k starts at 1.
For k=1: c=8, b=3*1 -1=2. So, a=1, b=2, c=8. Now, check if they are distinct: yes, all are different. Now, compute 3a + 2b + c = 3*1 + 2*2 + 8 = 3 + 4 + 8 = 15.
For k=2: c=16, b=5. Then, 3a + 2b + c = 3 + 10 + 16 = 29, which is larger than 15.
So, for a=1, the smallest value is 15.
Next, a=2:
Then, b = (3c - 8)/4. Since b must be a natural number, (3c - 8) must be divisible by 4. So, 3c ≡ 8 mod 4 => 3c ≡ 0 mod 4. Since 3 ≡ 3 mod 4, c must be ≡ 0 mod 4. So, c=4m, where m is a natural number.
Then, b = (3*4m - 8)/4 = (12m - 8)/4 = 3m - 2. So, b=3m -2 must be ≥1, so m ≥1.
For m=1: c=4, b=1. But a=2, b=1, c=4. All distinct. Compute 3a + 2b + c = 6 + 2 + 4 = 12.
For m=2: c=8, b=4. Then, 3a + 2b + c = 6 + 8 + 8 = 22, which is larger than 12.
So, for a=2, the smallest value is 12.
Next, a=4:
Then, b = (3c - 8)/2. Since b must be a natural number, (3c - 8) must be even. So,