Can you Solve Mathematical Problem from Harvard University Entrance Exam ?

[@KyyTyy]

42=32+8+2 2¹+2³+2⁵ (2^p)=2¹ (2^q)=2³ (2^r)=2⁵ p=1, q=3,r=5

[@oahuhawaii2141]

There's no domain restriction on p, q, and r. 2^p + 2^q + 2^r = 42 , p < q < r Let p = 0 and q = 1 . 2⁰ + 2¹ + 2ʳ = 42 2ʳ = 39 r = log₂(39) One of an infinite number of solutions: p = 0 , q = 1 , r = log(39)/log(2) ≈ 5.2854022188622... Alternate solution: 42 = 2Ah = 101010b = 2¹ + 2³ + 2⁵ p = 1 , q = 3 , r = 5

[@oahuhawaii2141]

I like this best: p = e q = π r = log₂(42-2^e-2^π) ≈ 4.73303627...

[@oahuhawaii2141]

I like this solution best: p = e q = π r = log₂(42 - 2^e - 2^π) = log(42 - 2^e - 2^π)/log(2) ≈ 4.7330...

[@karvindmishra09]

p=1, q=3, r=5

[@michaeledwards2251]

A Douglas Adams "Hitch Hikers" guide to the Galaxy reference. The computer gives the answer "42" to the question "life, the universe, and everything". Given computers today use binary, the question assumes binary is still used, and asks the question, How would the computer represent 42 ?

[@oahuhawaii2141]

Some solid-state memory storage devices save data in di-bit form; each memory cell has 4 states. That's base 4, so 42 will be encoded as 222₄ . [My circuit used an MCU with internal EEPROM based on Fowler-Nordheim tunneling di-bit memory technology.] BTW, the story continues with the people designing an even more powerful computer to figure out what was the question that resulted in 42. The answer is "What is 6 times 9?" Note that: 6*9 = 54 = 42₁₃ . [I can see base 6 or 12 being used, but not base 13.]

[@michaeledwards2251]

@@oahuhawaii2141 The minimum number of states to represent numbers occurs with base 3 : too awkward to work with. Both base 2 and 4 are equally efficient. Base 13 covers the "Bakers Dozen" as a unit digit : binary 1101 : base 3 as 111 : base 13 makes sense as a triple digit base 3 number.

[@oahuhawaii2141]

@@michaeledwards2251: A Bakers Dozen is 13, which is 10 in base 13. Not a unit digit. Numbers 0 to 12 can be encoded with 0 to 9, A-C .

[@에스피-z2g]

Solution by insight 2^p+2^q+2^r=42 Among 2,4,8,16,32 2+8+32=42 The only possible combination. p=1, q=3,r=5

[@superacademy247]

Thanks for sharing your insight 💯💕✅💪

[@oahuhawaii2141]

Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions: p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622... p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372... p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423... p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

[@oahuhawaii2141]

Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions: p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622... p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372... p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423... p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

[@oahuhawaii2141]

Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions: p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022188622... p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316186372... p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250014423... p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.7330362716713...

[@oahuhawaii2141]

Untrue. The requirement is p < q < r , but they can be real numbers. Here are 4 of an infinite number of valid solutions: p = 0 , q = 1 , r = log₂(39) = log(39)/log(2) ≈ 5.2854022... p = 2 , q = 4 , r = log₂(22) = log(22)/log(2) ≈ 4.4594316... p = 0 , r = 5 , q = log₂(9) = log(9)/log(2) ≈ 3.1699250... p = e , q = π , r = log₂(42-2^e-2^π) = log(42--2^e-2^π)/log(2) ≈ 4.733036...

[@sidharthsehrawat6423]

1 3 5

[@key_board_x]

2^(p) + 2^(q) + 2^(r) = 42 2^(p) + 2^(q + p - p) + 2^(r + p - p) = 42 2^(p) + 2^(p + q - p) + 2^(p + r - p) = 42 2^(p) + [2^(p) * 2^(q - p)] + [2^(p) * 2^(r - p)] = 42 2^(p).[1 + 2^(q - p) + 2^(r - p)] = 42 ← there is an odd number in the second bracket 2^(p).[1 + 2^(q - p) + 2^(r - p)] = 2 * 21 2^(p).[1 + 2^(q - p) + 2^(r - p)] = 2^(1) * 21 → you can deduce that: 2^(p) = 2^(1) → p = 1 [1 + 2^(q - p) + 2^(r - p)] = 21 2^(q - p) + 2^(r - p) = 20 → recall: p = 1 2^(q - 1) + 2^(r - 1) = 20 [2^(q) * 2^(- 1)] + [2^(r) * 2^(- 1)] = 20 → recall: x^(- a) = 1/x^(a) → then: 2^(- 1) = 1/2^(1) = 1/2 [2^(q) * 1/2] + [2^(r) * 1/2] = 20 (1/2).[2^(q) + 2^(r)] = 20 2^(q) + 2^(r) = 40 2^(q) + 2^(r + q - q) = 40 2^(q) + 2^(q + r - q) = 40 2^(q) + [2^(q) * 2^(r - q)] = 40 2^(q).[1 + 2^(r - q)] = 40 ← there is an odd number in the second bracket 2^(q).[1 + 2^(r - q)] = 8 * 5 2^(q).[1 + 2^(r - q)] = 2^(3) * 5 → you can deduce that: 2^(q) = 2^(3) → q = 3 [1 + 2^(r - q)] = 5 2^(r - q) = 4 2^(r - q) = 2^(2) r - q = 2 r = 2 + q → recall: q = 3 → r = 5 Summarize p = 1 q = 3 r = 5

[@superacademy247]

Awesome 💯😎👍

[@kareolaussen819]

This is a question of finding the binary representation of 42. Take r>q>p and define r to be the largest number for which 2^r ≤ 42. Subtract and repeat the procedure for the remainder (here 10) until the remainder is zero. Here we find 42 = 32 + 8 + 2 = 2^5 + 2^3 + 2^1. I.e, r, q, p = 5, 3, 1.

[@oahuhawaii2141]

No other permutation because of the relational condition.

[@kareolaussen819]

@@oahuhawaii2141 You a right! I wrote the first version of my comment before I read that condition, and forgot to change the last part of my comment.

[@oahuhawaii2141]

@@kareolaussen819: The only restriction is p < q < r , so I chose real numbers. I like this solution best: p = e q = π r = log₂(42 - 2^e - 2^π) = log(42 - 2^e - 2^π)/log(2) ≈ 4.7330...

[@TheNizzer]

32,16,8,4,2,1,… (0 removed) Which 3 sum to 42? 32,8,2 P=1, q=3, r=5. This is supposed to be difficult? Given non integer powers of 2 are irrational….

[@oahuhawaii2141]

Untrue, as 2 raised to a negative integer is a fraction, which isn't irrational. Example: 2⁻⁵ = 1/32 = 0.03125 .

[@TheNizzer]

@ and three negative powers add up to more than 1? They aren’t under consideration, for obvious reasons.

[@oahuhawaii2141]

​@@TheNizzer: You wrote "non integer powers of 2 are irrational". My example shows your statement is incorrect. BTW, in your list of numbers, each corresponds to powers of 2, with exponents from 5 down to 0, except the last entry, where you have 2ⁿ = 0 . No such n exists.

[@TheNizzer]

@@oahuhawaii2141 Are you claiming -1, -2 … are non integer? Or that eg 2^0.5 is rational? Or eg 2^-0.5 is in some way rational? Can you advise a non integer power that is rational as a counter example?

[@oahuhawaii2141]

@@TheNizzer: It's possible to get an integer result by raising 2 to an irrational number: 2^(log(3)/log(2)) = 3 .

[@saltydog584]

Just write it as a binary number. 32=10101 then convert back to base 10. 10101=32 + 8 + 2 = 2^5 + 2^3+2^1so P is 5, Q is 3 and r =1 (only works when P,Q&R are whole numbers)

[@pijusuaac.chattopadhakjm2676]

How is 32(base 10)=10101(base 2)?

[@oahuhawaii2141]

@pijusuaac.chattopadhakjm2676: OP is very sloppy. I would show the conversion as: 42 = 2Ah = 101010b = 2⁵ + 2³ + 2¹

[@saltydog584]

@@pijusuaac.chattopadhakjm2676 Thanks, I've corrected my typo.