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Set of 3 – Part 2

Posted by Karl Sharman on February 1, 2011 – 3:00 pm

1. There is one number less than 1000 that gives a remainder of 1 when it is divided by 2, 3, 4, 5, 6, 7 or 8. What is that number?

2. I have five letters and five addressed envelopes. If I place the letters in the envelopes at random, what are the chances that only four letters are in their correct envelopes? Chris, our resident probabilities expert is definitely going to struggle with this one! ;-)

3. A driver sets out on a 20-mile trip. When he has gone halfway he finds he has averaged 25 mph. At what speed must he travel the rest of the way to make his overall average speed for the trip 40 mph?

This post is under “Tom” and has 13 respond so far.
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13 Responds so far- Add one»

  1. 1. Chris Said:

    eye nose

  2. 2. Chris Said:

    2. I did a more general version of that about a year ago. It is harder than it looks (if I’m remembering it right).

  3. 3. Bekki Said:

    question 1: LCM(2,3,4,5,6,7,8)+1 = 840+1 = 841

    question 2: if 4 of them are in the right envelope, then there is only 1 in the wrong envelope. This is impossible so the probability is 0.
    let’s pretend that letter #1 is in the wrong envelope. That means that there’s a different letter in envelope #1. That would automatically make 2 wrong letter/envelope combinations.

  4. 4. mike Said:

    @ bekki
    lol, exactly what i was thinking about #2. It’s impossible to get only 4/5 correct.

  5. 5. Dual Aspect Said:


    If he has averaged 25mph over a 10 mile distance then he has taken 24 mins to travel that first 10 miles.

    To average 40mph over the full 20 miles he must do the whole journey in 30 mins.

    This means he must travel the second half of the journey in just 6 mins.

    10 miles in 6 mins = 120mph

  6. 6. cazayoux Said:

    10 miles in 6 mins = 120mph?

  7. 7. Chris Said:

    2. I fell for that one :(

    In revenge, I’ve posted “Misdirected” to show the conclusion that I’d jumped to.

  8. 8. Dual Aspect Said:

    Ooops, It was late!

    That’s my defence your honour, and I’m sticking to it.

    But I would like to change my plea to 100mph.

  9. 9. DP Said:

    i agree Dual Aspect, 100 mph it is.
    I also agree with Bekki on the first 2. #1 is kind of like that mokey/coconut problem a while back (only simplified), and i don’t think #2 can happen..exactly 4 correct envelopes isn’t one of the possibilities. that is unless you lost on of the letters under the couch or something.

  10. 10. John Said:

    Solution to the first question is 841.

    First we multiply all the prime numbers: 2*3*5*7(=210). 210 is divisible 2,3,5,7.

    A number is divisible by 6 if and only if it is divisible by 2 and 3. Since 210 is divisible by 2 and 3, it implies 210 is also divisible by 6. Hence 210 is divisible by 2,3,5,6,7.

    Multiples of 4 are obtained by 4*x or 2*2*x. Now 210=2*(3*5*7), hence we multiply 210 by 2 to get 420=2*2*(3*5*7) which is a multiple of 4. Hence 420 is divisible by 2,3,4,5,6,7.

    Similarly, Multiples of 8 are obtained by 8*x or 2*2*2*x. Now 420=2*2*(3*5*7), hence we multiply 420 by 2 to get 840=2*2*2*(3*5*7) which is a multiple of 8.

    Hence 840 is divisible by 2,3,4,5,6,7 and 8.

    Since we have 840, which gives a remainder of 0 when it is divided by 2,3,4,5,6,7 or 8, by adding 1, we get a number which gives a remainder of 1 when it is divided by 2,3,4,5,6,7 or 8.

    Hence 841 gives a remainder of 1 when it is divided by 2,3,4,5,6,7 or 8.

  11. 11. Chris Said:

    To find the LCM, either enter LCM(2,3,4,5,6,7,8) into wolframalpha or:

    Break each number into its prime factors, and multiply the highest powered factors in the set together. So
    2,3,4,5,6,7,8 => 2¹,3¹,2²,5¹,2¹*3¹,7¹,2³ => 2³*3¹*5¹*7¹ = 840

  12. 12. Karl Sharman Said:

    Nothing to add here – the correct answers are:

    841 – Bekki

    Nil – if four are correct, then the fifth is also… – Bekki

    100mph! – DualAspect.

  13. 13. Chris Said:

    If and only if we have two numbers m,n, then
    |m*n| = abs(m*n) = lcm(m,n)*gcd(m,n)

    lcm and gcd > 0, regardles of the signs of m and n.

    There is no such simple formula when three or more numbers are involved.

    lcm and gcd obey handy rules e.g. lcm(a,lcm(b,c)) =

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