2 practice questions off the AMC!!!! (1 Viewer)

[lookoutastroboy]

HEY GUYS,

i need help solving these two problems and if you could supply ur answers and solutions to each one, that'd be greatly appreciated.

9. What is the maximumsize of a subset,​

P, of {1, 2, 3, . . . , 50} with the property that no pair of distinct elements of P​

has a sum divisible by 7?
(A) 21 (B) 22 (C) 23 (D) 24 (E) 25
​

10.​

​

How many three-digit numbers from 100 to 999 inclusive
have one digit which is the average of the other two?​

(A) 121 (B) 117 (C) 112 (D) 115 (E) 105

Thanks in advance, lookoutastroboy
(P.S. if you could tell me the specific mathematical methods to approach each question, that'd also be another great help!!! )

 

[u-borat]

i'm pretty sure you're meant to do the AMC yourself with no outside help.

 

[lookoutastroboy]

i'm pretty sure you're meant to do the AMC yourself with no outside help.

these are practice questions, not the competition ones, they're meant to be used for practice purposes so I dont see why I can't get help in these questions

 

[u-borat]

sorry, my bad.

 

[lookoutastroboy]

sorry, my bad.

no probs,

ummm.. u-borat, any chance of you helping me solve these problems since im assuming your pretty smart as your doing 4U Maths and English =)

 

[lolokay]

10.

Spoiler

A) 121
let: the middle digit be the average, the other digits not be equal to each other, an the last digit to not be a 0 (ie no 0's in the number)
in this case, you can see that the first and last digit are either both odd or both even
now, I added a case that they were also not equal so you have 4*3 = 12 (both even) + 5*4 = 20 (both odd) ways of doing this, ie 32 ways
Now, you have the above conditions but the average can be either of the 3 digit then you triple this value, giving a total of 96
Now, say all the digits are equal. There is 9 such numbers, giving a total of 105
Now, let the last digit be 0, with the second digit the average. You can see you have four ways of doing this - but for each of these 4 numbers there is 4 ways of ordering them such that the 0 does not become the first digit, so a total of 16 ways. This brings the total to 121, and you have covered all possibilities

I don't understand what question 9 means

 

[sicmacao]

HEY GUYS,

i need help solving these two problems and if you could supply ur answers and solutions to each one, that'd be greatly appreciated.

9. What is the maximumsize of a subset,​

P, of {1, 2, 3, . . . , 50} with the property that no pair of distinct elements of P​

has a sum divisible by 7?
(A) 21 (B) 22 (C) 23 (D) 24 (E) 25
​

10.​

​

How many three-digit numbers from 100 to 999 inclusive
have one digit which is the average of the other two?​

(A) 121 (B) 117 (C) 112 (D) 115 (E) 105

Thanks in advance, lookoutastroboy
(P.S. if you could tell me the specific mathematical methods to approach each question, that'd also be another great help!!! )

Question 9 Ans: C 23

The question asks for a set that contains numbers from 1 to 50 inclusive such that the sum of any pair of numbers are not divisble by 7. How many numbers would such a set contains if it should contain as many numbers as possible.

I will list the numbers to illustrate the idea.
1, 2, 3, 7
1*7+1, 1*7+2, 1*7+3
2*7+1, 2*7+2, 2*7+3
3*7+1, 3*7+2, 3*7+3
4*7+1, 4*7+2, 4*7+3
5*7+1, 5*7+2, 5*7+3
6*7+1, 6*7+2, 6*7+3
7*7+1
Hence the largest subset must contain 23 numbers. The subset I've shown is not unique. You may find another subset with different numbers as an exercise.

 

[lolokay]

Question 9 Ans: C 23

The question asks for a set that contains numbers from 1 to 50 inclusive such that the sum of any pair of numbers are not divisble by 7. How many numbers would such a set contains if it should contain as many numbers as possible.

oh I get it

my method would be:
convert your numbers into base 7 - ie number only have digits 0-6
the maximum value you can have is 50, which is 71 in base 7.
now, a set will contain no pairs that sum to a number divisible 7 if none of the digits in the units of the two numbers sum to 7, or to 0
so you have
1,2,3
11,12,13
...
61,62,63
71
= 7 sets of 3 + 1 = 22

you can also have one number ending in 0, eg 10, bringing the total to 23

I think there's a total of 28 such 23-element sets