Maize counted her dime.when she put them in groups of 4 ,she had two dimes left over.when she put them in group of 5,she had one left over.what is the smallest number of dimes she could have.if she has more than 10?
Joshua gave warren a birthday present,how much ribbon did he need to go around the present and make the bow?the bow took 12 inches by itself.
You have to work with piles = x
4x + 2 = 5x +1
x = 3 if you substitute you have 14 dimes
3 piles of 4 = 12 plus 2 = 14
3 piles of
Sorry.. I didn't mean to post that.
the answer gives us x =1 not 3
and we get to 6 dimes, but your problem is talking about more than 10 dimes.
The one pile works because you have
4(1)+ 2 = 5(1) + 1
6 =6 .
Now, we have to find a multiple that will work...
Do you need more help from here?
"when she put them in groups of 4 ,she had two dimes left over"
---> 6 10 14 18 22 26 30 34 38 42 46 50 ...
"when she put them in group of 5,she had one left over"
----> 6 11 16 21 26 31 36 41 46 51 ...
ahhh, looks like 46
actually, I just noticed that 26 works first.
To find the smallest number of dimes Maize could have, we need to find a number that satisfies the given conditions: when Maize puts the dimes in groups of 4, she has two leftovers, and when she puts them in groups of 5, she has one leftover.
Let's start by finding a number greater than 10 that satisfies the first condition (having two leftovers when divided into groups of 4). We can use trial and error by starting with numbers greater than 10 and counting by intervals of 4.
Starting from 12, when we divide it into groups of 4, we get 3 groups of 4 dimes, which leaves 0 dimes as leftovers. So, 12 is not the answer.
Next, let's try 13. When we divide 13 into groups of 4, we get 3 groups of 4 dimes, leaving 1 dime as a leftover. So, 13 could be a possible answer.
Now let's check the second condition, which states that when Maize puts the dimes in groups of 5, she has one leftover. We will test if 13 meets this condition.
When we divide 13 into groups of 5, we get 2 groups of 5 dimes, leaving 3 dimes as leftovers. Since 13 does not satisfy the second condition, it is not the answer.
Let's continue with the next number, 14. When we divide 14 into groups of 4, we get 3 groups of 4 dimes, leaving 2 dimes as leftovers. So, 14 could be a possible answer.
When we divide 14 into groups of 5, we get 2 groups of 5 dimes, leaving 4 dimes as leftovers. Since 14 does not satisfy the second condition, it is not the answer.
Let's try the next number, 15. When we divide 15 into groups of 4, we get 3 groups of 4 dimes, leaving 3 dimes as leftovers. So, 15 could be a possible answer.
When we divide 15 into groups of 5, we get 3 groups of 5 dimes, leaving 0 dimes as leftovers. Since 15 does not satisfy the second condition, it is not the answer.
Continuing this process, we find that 16 satisfies both conditions:
When we divide 16 into groups of 4, we get 4 groups of 4 dimes, leaving 0 dimes as leftovers.
When we divide 16 into groups of 5, we get 3 groups of 5 dimes, leaving 1 dime as a leftover.
Therefore, the smallest number of dimes Maize could have, given that she has more than 10, is 16.