Difficult Problems 2

nippon276

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While some of you have read my first post, there are still some problems that give me confusion:
1. If n is an integer such that 101< n <118 , for how many distinct values of n is the expression 11(16n )−1 a prime?

I know the answer is 0 from the answer key, but I would like an explanation. Originally I thought thatt it was 0 because of multiplication (then I found the -1). What logic would one use to check (since my calculator results in overflow error).

2. Let N be a three-digit number (with a non-zero hundreds digit) such that the sum of the digits of N is eight. If N2 is divided by 9, the remainder is 1. How many distinct possibilities exist for the value of N ?

I've gotten so far as to see 8 possible digits (0-7) and then was lost.

3. An ordinary clock with a minute hand and hour hand now indicates it is 12:00 P. M. Noon. The minute and hour hands are together at the usual dial positions, but the clock is actually running slow. When the minute and hour hands are first together after 3:00 pm, the clock has actually been running for 210 minutes. Find the number of true minutes lost by the clock during the time from 12:00 P. M. Noon on the clock to the first time after 3:00 P. M. that the minute and hour hands came together. Express your answer as an improper fraction reduced to lowest terms.

:confused: (answer is 150/11)

4. Emily travels at a rate of 4k mph. for 4 miles, 5k mph. for 5 miles, etc., and finally at the rate of wk mph. for w miles. For what value of w will the total distance traveled divided by the total time it took (average rate) be 12 times her starting rate?

Also :confused:

5. If √(x)=√(600+√(600+√(600+√(600+...)))) (an infinitely nested radical), find the value of x.

Ok if the x wasn't in a radical I see how one could solve it. You could square both sides (x2=600+x (the last x is from substitution)). The answer would be 25 (the -24 wouldn't work) if x had not been a radicand. I see that in the answer key it says 625...
 
5. If √(x)=√(600+√(600+√(600+√(600+...)))) (an infinitely nested radical), find the value of x.

Ok if the x wasn't in a radical I see how one could solve it. You could square both sides (x2=600+x (the last x is from substitution)). The answer would be 25 (the -24 wouldn't work) if x had not been a radicand. I see that in the answer key it says 625...
You could 1st square both sides and then get x= 600 + sqrt(x). You can get a quadratic equation from this. Just reject any solution with x<600.
Jomo
 
2. Let N be a three-digit number (with a non-zero hundreds digit) such that the sum of the digits of N is eight. If N2 is divided by 9, the remainder is 1. How many distinct possibilities exist for the value of N ?
Here is a hint. If the sum of the digits in N add up to 8 then any two digits that N can be differ by a multiple of 9.
Good luck,
Jomo
 
You could 1st square both sides and then get x= 600 + sqrt(x). You can get a quadratic equation from this. Just reject any solution with x<600.
Wow... I can't believe I didn't see that. Thank you for opening up my eyes

Here is a hint. If the sum of the digits in N add up to 8 then any two digits that N can be differ by a multiple of 9.
Could I get another hint? :smile:
 
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1. If n is an integer such that 101< n <118 , for how many distinct values of n is the expression 11(16n )−1 a prime?

We have
11 * 16n - 1 = 11 * (16n -1 ) + 10 = 11 * (42n -1) + 10 = 11 ( 4n -1) (4n+1) + 10
Now suppose n is odd. Then 4n ends in 4 and (4n + 1) ends in 5 which means it is divisible by 5. But then so is 10. So suppose n is even. The 4n ends in 6 and (4n - 1) ends in 5 which means it is divisible by 5. But then so is 10.

2. Let N be a three-digit number (with a non-zero hundreds digit) such that the sum of the digits of N is eight. If N2 is divided by 9, the remainder is 1. How many distinct possibilities exist for the value of N ?

Let
N = 100 a + 10 b + c
where a is positive and b and c are non-negative. Then
N2 = 104 a2 + 102 b2 + c2 + 2 (103 a b + 102 a c + 10 b c)
Let s = the sum of the digits of N and S the sum of the digits of N2. Then looking at the equation for N2 we see that
S = s2
So, if s is 8 then S is 64 and the fact that N2 leaves a remainder of 1 when divided by 9 is no extra information. So start counting:
a=1: b,c = 0, 7; 1, 6; 2, 5; 3, 4; 4, 3; ... a total of 8
a=2: b,c = 0, 6; 1, 5; ... a total of 7
a total of 6
5
4
...
a=8: b,c = 0,0 a total of 1
so 1 + 2 + ... + 8 = 32

3. An ordinary clock with a minute hand and hour hand now indicates it is 12:00 P. M. Noon. The minute and hour hands are together at the usual dial positions, but the clock is actually running slow. When the minute and hour hands are first together after 3:00 pm, the clock has actually been running for 210 minutes. Find the number of true minutes lost by the clock during the time from 12:00 P. M. Noon on the clock to the first time after 3:00 P. M. that the minute and hour hands came together. Express your answer as an improper fraction reduced to lowest terms.
Need to assume hands only have integer minute positions (or, for more accuracy, integer second positions). The hour hand goes from 3 to 4 in one hour, i.e. travels at 5 min/hr. The minute hand travels 60 min/hr. So, starting at 3, we have
M = 0 + 60 t
H = 15 + 5 t
and they are the same at 55 t = 15 or t = 3/11 hr = 180 / 11 min or total indicated elapse time since noon of 3 hours and 180/11 min which is 2160/11 min. Actual time was 210 min or 2310/11 min. Thus time lost was (2310-2160) = 150/11 min


4. Emily travels at a rate of 4k mph. for 4 miles, 5k mph. for 5 miles, etc., and finally at the rate of wk mph. for w miles. For what value of w will the total distance traveled divided by the total time it took (average rate) be 12 times her starting rate?
t = miles/rate
D = 4 + 5 + 6 + ... + w = w (w+1) / 2 - 6
T = (1 + 1 + 1 + ... + 1)/k = (w-3)/k
D/T = 12 * 4

A quadratic in w. Don't forget the k which stands for 1000 I believe.
 
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or t = 3/11 hr = 180 / 11 min or total indicated elapse time since noon of 3 hours and 180/11 min which is 2160/11 min. Actual time was 210 min or 2310/11 min. Thus time lost was (2310 - 2160) = 150/11 min
Sorry—you kind of lost me here.

t = miles/rate
D = 4 + 5 + 6 + ... + w = w (w+1) / 2 - 6
T = (1 + 1 + 1 + ... + 1)/k = (w-3)/k
D/T = 12 * 4

A quadratic in w. Don't forget the k which stands for 1000 I believe.
I understand how you got D and I but I was unsure on how to solve from there. The answer is 92.
 
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Why isn't the k used after (w-3)?

I have no idea, it disappeared from the problem as though it were 1 rather than the 'normal' 1000.

Also, notice
w (w + 1) / 2 - 6 = (w2 + w - 12) / 2 = (w-3) (w + 4) / 2
so
[w (w + 1) /2 - 6] / (w-3) = (w + 4) / 2 = 48 or
w = 96 - 4 = 92.
 
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