A very tricky Question from Stanford University Admission Exam

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Комментарии : 8

@michaeledwards2251 19 часов назад

Convert 42 base 10 to 101010 base 2, or 2 power 5, 3, and 1 respectively, as used by all chip based arithmetic. It is one of the most common calculations performed globally. The answer in base 10 is r = 5, q = 3 and p = 1.

@dan-florinchereches4892 19 часов назад

I see this solution. I wonder if the problem is asking for integer solutions. If so this is the most down to earth way, otherwise we get an infinity of solutions based on various conditions I am considering a simpler case. Let Z=2^x+2^y X

@dr.walterstadler1837 22 часа назад

No need to create a doctorate job out of this. 2^n is all known and to combine 32 + 8 +2 = 42, so it is 5, 3, 1.

@KyyTyy 2 часа назад

2^p+2^q+2^r=42 2⁶=64 2⁵=32 2⁴=16 2³=8 2²=4 2¹=2 2⁰=1 2⁵=32 2³=8 2¹=2 2⁵+2³+2¹=42 P=5, q=3 r=1

@saurabhsrivastava4140 6 часов назад

Break 42 in set of 2 power number Set (2,4,8,16,32) And above three number and select the number whose sum is 42 (2,8,32) 2=2^1, 8=2^3, 32=2^5 Compair it with given equation So, we get p=1, q=3,r=5 answer

@rogerphelps9939 17 часов назад

Just write down the bi nary representation of 42 and the answer is obvious. That can be done in one line.

@key_board_x 22 часа назад

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