Some problems which I am stuck at? Any help? I beg anyone...

These questions are to be solved without a calculator:

1) Three different non-zero digits are used to form six different 3 digit numbers. The sum of 5 of them is 3321. What is the 6th number?

2) How many pairs \((a, b)\) of positive integers are there such that \(a\) and \(b\) are factors of \(6^6\) and \(a\) is a factor of \(b\)?

3) All digits of the positive integer \(N \)are either \(0\) or \(1\). The remainder after dividing\( N\) by \(37\) is \(18\). What is the smallest number of times that the digit \(1\) can appear in \(N\)?

4) In how many ways can the numbers 1, 2, 3, 4, 5, 6 be arranged in a row so that the product of any 2 adjacent numbers is even? Choices are: 64 or 72 or 120 or 144 or 720

5) A hockey game between two teams is 'relatively close' if the number of goals scored by the two teams differs by more than two. In how many can the first 12 goals of a game is scored if the game is "relatively close"?

6) The 4 digit number \(pqrs\) has the property \(pqrs \times srqp\). If \(p = 2\) what is the value of the 3 digit number \(qrs\)?

7) If \(x^2 = x + 3\), then what is the value of \(x^3\)? Choices are: A) x + 6 B) 2x + 6 C) 3x + 9 D) 4x + 3 E) 27x + 9

Please answer these questions with complete steps in the comment section below. I thank everyone who tries from the bottom of my heart.

EDITED!

Few more questions:

Q8) A floor tile has the shape of a regular polygon. If the tile is removed from the floor and rotated 50° it will fit back exactly into its original place in the floor. The least number of sides this polygon can have is: A) 8 B) 24 C) 25 D) 30 E)36

Q9) On my car, a particular brand of tyre lasts 40 000 kilometres on a front wheel or 60 000 km on a rear wheel. By interchanging with the front and rear tyres the greatest distance in km I can get from a set of 4 tyres of these is:

A) 52000 B) 50 000 C) 48 000 D) 40 000 E) 44 000

Q10) Each face of a solid cube is divided into 4 as indicated in the diagram. Starting from vertex P, paths can be travelled to vertex Q along connected line segments. If each movement along the path takes one closer to Q, find the number of possible paths from P to Q.

Note by Syed Hamza Khalid
3 months, 1 week ago

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Q9) The answer is the harmonic mean. \(\frac{2}{\frac{1}{60000}+\frac{1}{40000}}=48000\). (24000 miles wears the back tires 60% but the front 40%)

Jeremy Galvagni - 3 months ago

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Why do we use the harmonic mean?

Syed Hamza Khalid - 3 months ago

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I don't usually recognize harmonic mean problems until after the fact so that didn't help me solve it. To me this is more like problems of the type:

\(A\) can do the job in 4 hours, \(B\) can do the job in 6. How long does it take them working together?

In this tire problem, they do the job twice (front and back)

Jeremy Galvagni - 3 months ago

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@Jeremy Galvagni 4.8 hrs? Using harmonic sequence...

Syed Hamza Khalid - 3 months ago

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Q6 :Perhaps you mean pqrs=4*srqp?(2178x4=8712)

X X - 3 months, 1 week ago

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Yes but how did you find that number without trial and error as calculator isn't allowed

Syed Hamza Khalid - 3 months, 1 week ago

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This is one of the problems on brilliant.You can view the solution here

X X - 3 months, 1 week ago

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@X X Thanks

Syed Hamza Khalid - 3 months, 1 week ago

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@Syed Hamza Khalid Can you help with the new questions?

Syed Hamza Khalid - 3 months ago

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1) Call the digits a,b,c. The sum of all six 3-digit numbers is 222(a+b+c). 222(a+b+c)-3321=the 6th number. Call it 100c+10b+a. A table of 222n-3321 show when n=18 yields 675 whose digits sum to 18. So the 6th number is 675.

2) \(6^{6}=2^{6}3^{6}\) which has \(7\cdot 7=49\) factors. So there are 49 possible values of b. You can use the same trick to find the number of a for each b. The total is \(1+2+3+4+5+6+7)+(2+4+...14)+...+(7+14+...+49) = 28\cdot28 = 784 \)

3) \(10^n\) has only three possible remainders: 1, 10, 26. You seek to make a sum of 18. 26*2-37=15 which is really close, just need three 1's: The smallest arrangement is 1101101 and the answer is 5.

4) The odds must be separated. Considering just even/odd there are 4 arrangements: oeoeoe, oeoeeo, oeeoeo, eoeoeo. There are 3!=6 ways to place the odds and 3!=6 ways to place the evens. \(4\cdot6\cdot6=144\).

5) Makes no sense. I think maybe you mean 'relatively close' means they never differ by 2 (or maybe at most 2) and you want the ways the game can remain relatively close with 12 goals scored.

6) Seems to be missing the property.

7) Multiply by \(x\) to get \(x^{3}=x^{2}+3x\) then sub in the given \(x^{2}=x+3\) and simplify to \(x^{3}=4x+3\)

Jeremy Galvagni - 3 months, 1 week ago

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Can you help with the new questions?

Syed Hamza Khalid - 3 months ago

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For Q2, how did you find it has 49 factors .

Syed Hamza Khalid - 3 months ago

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To count the number of factors a number has, look at its prime factorization. Then add 1 to each exponent and multiply them together.

For example \(360=2^{3}3^{2}5^{1}\) and has \(4 \cdot 3 \cdot 2 = 24\) factors.

\(6^{6}=(2 \cdot 3)^{6} = 2^{6}3^{6}\) and has \(7 \cdot 7 = 49\) factors.

Jeremy Galvagni - 3 months ago

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@Jeremy Galvagni Can you tell me what you did after finding the factors?

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid The factors of \(2^63^6\) are in the form of

\(2^{m-1}3^{n-1}\) (\(m\in\{1,2,3,4,5,6,7\},n\in\{1,2,3,4,5,6,7\}\)), and \(2^m3^n\) has \(mn\) factors.

The sum of all the possibilities of \(m\) is \(1+2+3+4+5+6+7\). The sum of all the possibilities of \(n\) is also \(1+2+3+4+5+6+7\).

So, the sum of the possibilities of \(mn\) equals \((1+2+3+4+5+6+7)(1+2+3+4+5+6+7)=784\)

X X - 3 months ago

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@X X Hey X X, I have some doubts:

  1. Why do we write the exponential in the form of \( m - 1 \text{ and } n - 1\) ?

  2. How do you say that the possibilities of \( m \) is a set of 1,2,3,4,5,6,7 ? Is there a kind of theorem or something?

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid

  1. Because we have to add one to the exponential, so I prefer to "subtract one" first. It doesn't really affect the answer.
If I write the exponential in the form of \(m\) and \(n\), then it will simply become

"The factors of \(2^63^6\) are in the form of

\(2^{m}3^{n}\) (\(m\in\{0,1,2,3,4,5,6\},n\in\{0,1,2,3,4,5,6\}\)), and \(2^m3^n\) has \((m+1)(n+1)\) factors.

The sum of all the possibilities of \(m+1\) is \(1+2+3+4+5+6+7\). The sum of all the possibilities of \(n+1\) is also \(1+2+3+4+5+6+7\).

So, the sum of the possibilities of \((m+1)(n+1)\) equals \((1+2+3+4+5+6+7)(1+2+3+4+5+6+7)=784\)"

2.Because \(m-1\) cannot be smaller than \(0\), or bigger than \(6\)(otherwise, it would not be a factor of \(2^{m-1}3^{n-1}\)), so it can only be \(0,1,2,3,4,5\) or \(6\), so \(m\) can only be \(1,2,3,4,5,6\) or \(7\)

X X - 3 months ago

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@X X Thanks a lot. Your answers are legendary and very beautiful with in-detail language.

Syed Hamza Khalid - 3 months ago

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@X X Thanks a lot. By the way are you really 14 years old? If so, your maths is amazing. I am also 14 years

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid Yes, I'm really 14. I love math,so I study math myself.

X X - 3 months ago

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@X X Which grade? Me; going to 10 in September

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid I'm going to 9.

X X - 3 months ago

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@Jeremy Galvagni Unbelievable! I never knew that. Thanks a lot

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid You can view this

The factors of \(2^63^6\) is in the form of \(2^m3^n\), so \(m\in\{0,1,2,3,4,5,6\},n\in\{0,1,2,3,4,5,6\}\), hence there are \(7\times7\) possibilities of \(2^m3^n\)

X X - 3 months ago

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@X X Thanks

Syed Hamza Khalid - 3 months ago

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Can you re-give your answers somewhat more detailed for a grade 9 student? please for Q1, Q2, and Q3 only

James Bacon - 3 months ago

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For (Q1) you mentioned, "a table of 222n -3321"; how am I going to get a table from as it is a non-calculator question

Syed Hamza Khalid - 3 months, 1 week ago

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Good question, but it really isn't that hard to make a table for these numbers without a calculator, even though I used one.

You can do 222*15=3330 in your head. 3330-3321=9 then just count by 222's. (15, 9) (16, 231) (17, 453) (18, 675)

Jeremy Galvagni - 3 months, 1 week ago

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@Jeremy Galvagni How does 18 yield to 675?

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid 222*18-3321=675

Jeremy Galvagni - 3 months ago

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@Jeremy Galvagni Yes, but why do you choose specifically 18, why not another number?

Syed Hamza Khalid - 3 months ago

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@Syed Hamza Khalid 18=(a+b+c) so you need the correct digit sum and 18 = 6+7+5

Jeremy Galvagni - 3 months ago

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@Jeremy Galvagni Wat about 765

Syed Hamza Khalid - 1 month, 2 weeks ago

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@Jeremy Galvagni Oh Okay thanks

Syed Hamza Khalid - 3 months ago

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Q1 If the sixth number is \(\overline{abc}\), then \(3321 + \overline{abc} = 222(a+b+c)\). The smallest number that can be added to \(3321\) to give a multiple of \(222\) is \(9\), but \(009\) does not contain \(3\) distinct digits. Other possibilities of \(\overline{abc}\) are \(9+222k\), in which case \(a+b+c = 15+k\). A but of trial and error leads you to \(k=3\) and \(\overline{abc} = 675\).

Mark Hennings - 3 months, 1 week ago

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Can you help with the new questions?

Syed Hamza Khalid - 3 months ago

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For Q1, call them a,b,c so one number is 100a + 10b + c, 100a + 10c + b etc. Add all 6 up to get 222(a + b + c). Think from here

Stephen Mellor - 3 months, 1 week ago

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I got stuck right from here... :(

Syed Hamza Khalid - 3 months, 1 week ago

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Q10) I'm pretty sure the answer is 54. Number every point with the sum of all paths that lead to it. Begin with P=1. Q ends up as 18+18+18=54 if I didn't make any mistakes.

Jeremy Galvagni - 3 months ago

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Q8) I would just try 360/50 but it gives 7.2. Instead, reduce 360/50 to 36/5. The tile has 36 sides. Rotating 50 degrees skips to the 5th.

Jeremy Galvagni - 3 months ago

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That's exactly what I thought

Syed Hamza Khalid - 3 months ago

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For Q4, to avoid consecutive odd digits, the even digits must be either in positions 2,4,6 or 1,3,5. There are 3! * 3! ways for each case so 72

(also don't undertand Q6)

Stephen Mellor - 3 months, 1 week ago

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Can you help with the new questions?

Syed Hamza Khalid - 3 months ago

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not quite. Evens can also be in 2,4,5 or 2,3,5 because they can be next to each other.

Jeremy Galvagni - 3 months, 1 week ago

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oh yeah

Stephen Mellor - 3 months, 1 week ago

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For Q7, multiply all by x and then sub in the value of x^2 from what is given

Stephen Mellor - 3 months, 1 week ago

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Can you help with the new questions?

Syed Hamza Khalid - 3 months ago

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