Recent content by Robert Houdart
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68068 as difference of two squares
i I guess you are right. (a, b) and (b, a) are repeated in my solution, taking pair (a,b) and (b,a) as a single pair yields 8 as the answer.- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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How many ways can 6 cards be chosen from a deck to have all suits present?
I guess 8682544 is the correct answer.- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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68068 as difference of two squares
Homework Statement In how many ways can 68068 be written as the difference of two squares?Homework EquationsThe Attempt at a Solution Let (x+a) * (x+a) -x*x =68068=2*2*7*11*13*17 a (2x+a) =2*2*7*11*13*17 As 2x+a is odd ⇒ a is even ∴a=2b 2b (2x+2b) =2*2*7*11*13*17 b (x+b) =7*11*13*17 x=...- Robert Houdart
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- Difference Squares
- Replies: 5
- Forum: Precalculus Mathematics Homework Help
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How many ways can 6 cards be chosen from a deck to have all suits present?
Let us take suits as A, B, C, D Now the different case which arises are 3 A 1 B 1 C 1D 1 A 3 B 1 C 1D 1 A 1 B 3C 1D 1 A 1 B 1C, 3 D 2 A 2 B 1C, 1D 2 A 1 B 2C 1D 2 A 1 B 1C 2 D 1 A 2 B 2C 1D 1 A 2 B 1C 2 D 1 A 1 B 2C 2 D Therefore required number of ways will be 13*13*13 (4* 13c3) + 13*13*6...- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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How many ways can 6 cards be chosen from a deck to have all suits present?
Homework Statement In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?Homework EquationsThe Attempt at a Solution 4 different cards can be chosen in 13*13*13*13 ways. Now we have to choose 2 remaining cards from 48 cards. This can be done in...- Robert Houdart
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- Combination Permutation
- Replies: 11
- Forum: Precalculus Mathematics Homework Help
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Question about arithmetic progressions
Yes, if d=0 is considered, nine more numbers are added to the solution set (1111 , 2222..., 9999) However, my question is, a series (x, x, x, x, x, x, x) be considered an AP (that is can an AP have common difference=0)- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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Question about arithmetic progressions
Homework Statement Of a 4 digit positive integer, the four digits form an Arithmetic progression from left to right. How many such 4 digit integers exist? 2. The attempt at a solution If d = 1, the integers are 1234, 2345, …, 6789. These 6 integers and their reverses satisfy the given...- Robert Houdart
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- Arithmetic
- Replies: 3
- Forum: Precalculus Mathematics Homework Help
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How Many Non-Confusing Codes Can Be Formed from Distinct Digits?
(16,61) (18,81) (19,91) (68 ,86) (89, 98) i think these are all (actually total number of numbers were 81 instead of 90 (digits must be distinct)- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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How Many Non-Confusing Codes Can Be Formed from Distinct Digits?
well , considering 69 and 96 total number pertains to 10- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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How Many Non-Confusing Codes Can Be Formed from Distinct Digits?
Mod note: This post was a response in a separate cross-posted thread.Ok, I think I got it. Since they are distinct numbers, the first digit can be chosen in 9 ways (except 0) while the second can also be chosen in 9 ways, making it 81 numbers instead of 90.. So am I right this time?- Robert Houdart
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- Forum: Precalculus Mathematics Homework Help
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How Many Non-Confusing Codes Can Be Formed from Distinct Digits?
Homework Statement An intelligence agency forms a code of two distinct digits selected from 0, 1 , 2…, 9, such that the first digit of the code is nonzero. The code, handwritten on a slip, can, however, potentially create confusion when read upside down - for example; the code 91 may appear as...- Robert Houdart
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- Confusion
- Replies: 7
- Forum: Precalculus Mathematics Homework Help