Math problems

[smile101]

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1. Stella has 3 quarters and 4 dimes. How many different amounts can she form with these 7 coins, using at least one coin in each amount?

The answer is 19, but how do you calculate it w/out having to list all the possibilities?

2. If N is any positive integer, how many consecutive integers following N are needed to insure that at least one of the integers is divisible by another positive integer m?

I think the answer is m, but the solution said not to count N, so it would be m-1...don't know why you wouldn't count N.

 

[vvvv]

1. Stella has 3 quarters and 4 dimes. How many different amounts can she form with these 7 coins, using at least one coin in each amount?

The answer is 19, but how do you calculate it w/out having to list all the possibilities?

2. If N is any positive integer, how many consecutive integers following N are needed to insure that at least one of the integers is divisible by another positive integer m?

I think the answer is m, but the solution said not to count N, so it would be m-1...don't know why you wouldn't count N.

1. I sure there's the way to do this faster but I came up with a solution like this.
0quart and or 1, or 2, or 3, or 4 dimes so 4 diff amounts
1quart and 0, 1, 2,3,4 dimes so 5 diff amounts
2quart same with above 5 diff amounts
3quart same with above 5 diff amounts
total is 19.

 

[happyasaclam88]

i think the answer must have made a mistake since there are only 12 possible answers if you have to use at least one of each coin.

 

[smile101]

But you have to use one of each amount, so 0 is not an option!!!

 

[Streetwolf]

1. Stella has 3 quarters and 4 dimes. How many different amounts can she form with these 7 coins, using at least one coin in each amount?

The answer is 19, but how do you calculate it w/out having to list all the possibilities?

2. If N is any positive integer, how many consecutive integers following N are needed to insure that at least one of the integers is divisible by another positive integer m?

I think the answer is m, but the solution said not to count N, so it would be m-1...don't know why you wouldn't count N.

1) Since there is no way to get 0-4 dimes to equal any amount of quarters, you don't have to worry about counting something twice. If you could use 5 dimes then you'd have a problem since 5 dimes = 2 quarters and they want DIFFERENT amounts (so you would have to count 5 dimes to be the same as 2 quarters).

Knowing that, just realize that your options for quarters are to use 0, 1, 2, or 3 and that your options for dimes are to use 0, 1, 2, 3, or 4. Since you have 4 options and 5 options, you get 4*5 = 20. Furthermore, you need to use at least 1 coin, thus you can't use the 0 and 0 option. So you get 19 total.

2) It would work if one thing is meant by the problem: the integer N counts. In other words if I choose N = 10 and m = 5, the answer says that you need 4 additional integers: 11, 12, 13, and 14. This only works if you can count N = 10 as being the number divisible by 5. The question asks how many integers AFTER N you would need to guarantee divisibility. If N = 11 and m = 5, you would need 4: 12, 13, 14, and 15. So m-1 only works if you can count the starting integer N.

The answer key, when it says don't count N, means don't count it when you are counting the consecutive integers. If you say 5 consecutive integers with N = 11, you start with 12.

 

[snooper92]

Streetwolf,

Doesn't the answer depend on what value of "m" is chosen? This is why that question confused me.

 

[Streetwolf]

Yes that's why the answer is 'm-1' because it depends on what 'm' is.

If m = 7 you would need no more than 7 consecutive numbers to be certain that one of them is divisible by 7. Just think about it. Of those 7 numbers, one of them will have remainder 1 when divided by 7, another will have remainder 2, ..., remainder 6, and remainder 0.

If you choose N = 17 that means you need to consider 18, 19, 20, 21, 22, and 23.

17 has a remainder of 3. 18 has a remainder of 4, 19 has 5, 20 has 6, 21 has 0 (this is divisible by 7), 22 has 1, and 23 has 2.