The nine digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are each used exactly once in writing three 3-digit numbers. In each of the three numbers, the middle digit is the largest. What is the smallest possible sum of these three numbers?

The priority is getting the lowest hundred's digits, then the lowest tens digits and then the lowest ones digits.

So The numbers should be \( 1 \_ \_, 2 \_ \_, 3 \_ \_ \). Next, we need to put the tens digits. We can't put a 4, because then we can't put anything in the ones place. But we can put a 5; putting a 4 in the ones place. Likewise, we can't place a 6, since both 5,4 are used, but we can put a 7. Like wise for 8,9. Putting them in place, we get \( 154,276,398 \) with a sum of \( \boxed{828} \)

ABCDE is an irregular pentagon with DC = DE and the angles at E and C are right angles. Angle DAB is 70 degrees and angle DBA is 60 degrees. Also, EA + BC = AB. Determine the size of the angle EDC in degrees.

Hi Calvin. Thanks for the response. Won't the grid only work for determining the number of unique products if exactly TWO numbers are chosen? How can you determine the number of unique products if three or more numbers are chosen and multiplied?

@Mark Mottian
–
oh ooops, did not see "or more". That makes it much easier then.

Claim: Each product can be uniquely written as \( 3^a \times 4^b \times5 ^c \times 6 ^ d \times 7 ^ e \), where \( 0 \leq a \leq 1, 0 \leq b \leq 2, 0 \leq c \leq 2, 0 \leq d \leq 1, 0 \leq e \leq 3 \).
This happens to work for the set of numbers, and it should be rigorously proved. It would not work if we had 2 3's and 2 6's.

So, there are \( 2 \times 3 \times 3 \times 2 \times 4 \) possible products.
The number of products of 0 terms is 1. The number of products of 1 term is 5. So subtract \( 1 +5 = 6 \) from the previous total.

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## Comments

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TopNewestQuestion 16Let

fbe a function satisfying:\(f(xy) = f(x)/y \)

for all positive real numbers x and y. If \(f(500) = 3\), what is the value of \(f(100)\)?

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Isn't this a one line solution?

\( f(5*100) = f(100)/5 \implies f(100) = 5*f(500) = \boxed{15} \)

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Hi Siddhartha. As usual, it's an absolute pleasure to discuss things with you. Yes, that's my exact solution!

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Question 7The nine digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are each used exactly once in writing three 3-digit numbers. In each of the three numbers, the middle digit is the largest. What is the smallest possible sum of these three numbers?

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The priority is getting the lowest hundred's digits, then the lowest tens digits and then the lowest ones digits.

So The numbers should be \( 1 \_ \_, 2 \_ \_, 3 \_ \_ \). Next, we need to put the tens digits. We can't put a 4, because then we can't put anything in the ones place. But we can put a 5; putting a 4 in the ones place. Likewise, we can't place a 6, since both 5,4 are used, but we can put a 7. Like wise for 8,9. Putting them in place, we get \( 154,276,398 \) with a sum of \( \boxed{828} \)

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Well done! Can you please help me solve the last two problems as well?

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Question 19ABCDE is an irregular pentagon with DC = DE and the angles at E and C are right angles. Angle DAB is 70 degrees and angle DBA is 60 degrees. Also, EA + BC = AB. Determine the size of the angle EDC in degrees.

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Question 20How many different numbers can be written as the product of two or more of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7 ?

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A simple way to approach this question is to write the numbers out in a grid, and then multiply them out while ignoring terms in the main diagonal.

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Hi Calvin. Thanks for the response. Won't the grid only work for determining the number of unique products if exactly TWO numbers are chosen? How can you determine the number of unique products if three or more numbers are chosen and multiplied?

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Claim: Each product can be uniquely written as \( 3^a \times 4^b \times5 ^c \times 6 ^ d \times 7 ^ e \), where \( 0 \leq a \leq 1, 0 \leq b \leq 2, 0 \leq c \leq 2, 0 \leq d \leq 1, 0 \leq e \leq 3 \). This happens to work for the set of numbers, and it should be rigorously proved. It would not work if we had 2 3's and 2 6's.

So, there are \( 2 \times 3 \times 3 \times 2 \times 4 \) possible products. The number of products of 0 terms is 1. The number of products of 1 term is 5. So subtract \( 1 +5 = 6 \) from the previous total.

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