The digit on hundreds place of a three-digit number is 2, the digit on tens place is 3 less than the digit on ones place. 6 times the product of the three digits is 20 less than this number. What is this number?
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$\begingroup$ The construction of the number allows only 7 values. 203, 214, 225, 236, 247, 258, 269. 6p-20 is even, so only 214, 236 and 258 remain. Checking them yields 236. $\endgroup$– Florian FJun 19, 2020 at 10:57
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3 Answers
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Although it looks more like a problem rather than a puzzle, I believe that it's somewhat ontopic. Here's a solution that doesn't involve any algebra:
Note that 20 plus 6 times the product of the digits is the number itself, which must be something in the 200s. The remaining digits (apart from 2) must be 36 (since the last digit must be even (the number is 20 + multiple of 12), but 14 or 58 are not divisible by 4 - of course, here we check that's the "tens" digit is 3 less than "units"). Let's check: 236 indeed is 20+6*2*3*6, so final answer is 236.
answered Jun 16, 2020 at 20:18
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After drafting some simple equations, the answer can quickly be found with some easy mental math
The number must be in the format: "2XY"
where X = Y - 3
and where the solution to 6 * 2 * Y * (Y - 3) equals "2XY" - 20
thus, the solution to multiplying the terms together must also end in the digit Y (because 20 ends in 0)
thus, we need a number, Y, such that 2Y^2 - 6Y ends in the digit Y (for easier mental math, we can leave out multiplying by 6, as this only concerns the ones place)
a quick mental try of all 10 digits reveals that only 0 and 6 fit this criteria
0 does not work with the rest of the stipulations, so 6 must be the answer
Y equals 6, and we can solve for X (X = Y - 3 = 3)
The answer then comes out to 236
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This falls pretty close to the "math problem, not a puzzle" territory, but let's try to solve it anyways.
Write everything into an equation (using T as the digit in the "tens" place):
$$ 200 + T\times10 + (T+3) - 20 = 6\times 2 \times T \times (T+3) $$
and solve it:
$$ 11T + 183 = 12(T^2+3T) = 12T^2 + 36T$$ $$ 12 T^2 + 25T - 183 = 0 $$ $$ (T-3)(12T+61) = 0$$ $$ T=3 \quad \text{or} \quad T=-\frac{61}{12} $$
So we have two candidates for T, but one of them is not an integer between 1-6, so we pick the suitable one for the final result of
236
